The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
The most recent photograph of Pluto taken by New Horizons that I've been able to find is this one, from September 2006. Since the probe is now at a little over 5 AU from the dwarf planet, I'm wondering why I've not been able to find more recent pictures. New Horizons took many photographs of Jupiter and its moons during a flyby in 2006, so it seems that they would have also planned to photograph Pluto and Charon during the approach, and not just during flyby in 2015. I've not done the math to calculate what kind of angular diameter Pluto would have from New Horizons' current location, but I have to imagine that any image taken from the probe right now would provide better detail than the embarrasingly-blurry Hubble composite (though I admittedly know little about the technical specs of the cameras on board). Evanh2008(talk|contribs) 00:08, 7 February 2013 (UTC)[reply]
Here's a photo of Pluto, taken by LORRI, released from January 24, 2008. There's no need to remain in the dark: the science payload description contains detailed descriptions of each camera on board. Right now, the spacecraft is in the interplanetary cruise phase, so it isn't actively taking photos very often. It's pretty easy to know what the photos would look like: LORRI, the highest-spatial-resolution camera, has a 20.8 cm aperture, so you can trivially compute the farthest distance before Pluto will resolve to anything more than a single pixel. (About 500 million miles from Pluto, which is ... still a little ways to go). Space is very big! Nimur (talk) 00:20, 7 February 2013 (UTC)[reply]
Using a camera entails risks of damage, so taking lots of low-res pics is a bad risk-versus-reward option. They might take a pic occasionally, just to ensure that the camera is still working, but that's it, 'til they get closer. StuRat (talk) 00:27, 7 February 2013 (UTC)[reply]
A single pixel? According to my research (which consists of watching detective shows on network TV), you can expand a single pixel into a high-resolution picture. Just tell the technician "enhance!"... (Note to the humor impaired: no, I am not being serious) --Guy Macon (talk) 06:14, 7 February 2013 (UTC)[reply]
The ones that annoy me the most are when the technician sitting at the keyboard apparently doesn't think of enhancing the picture until the non-technical guy looking over his shoulder says "Can you enhance that?"..."Oh! Yes! I can! I'm glad you suggested that!"...Star Trek is frequently guilty of this one. SteveBaker (talk) 14:10, 7 February 2013 (UTC)[reply]
And, in the audio field, you had to love the original Star Trek episode where they detected the presence of an extra person on the ship by turning the ship's microphones way up and listening for heartbeats. I was worried that Spock would fart and permanently deafen them all. StuRat (talk) 16:31, 7 February 2013 (UTC) [reply]
Did you notice that, while their pulses were at all sorts of phases with respect to each other, the whole damm lot of them had the exact same clock-steady pulse rate? It doesn't happen that way in practice. In any group of healthy people, you get rest pulse rates anywhere from below 60 to as much as 75 or even more. Surely someone in the ship actually getting some physical exercise would have a varied rate? And some people past about age 40 or so may have the odd ectopic beat now and then. Wickwack 58.167.247.139 (talk) 00:35, 8 February 2013 (UTC) [reply]
Ack. That's why I don't like most Star Trek episodes. Their 23rd century is so clean and shiny that their Schroedinger's cat is 100% alive and even their Black Holes are actually only an unusually dark shade of gray.
And Spock doesn't fart. He simply holds it back till it comes out the other end as pseudoscience. - ¡Ouch! (hurt me / more pain) 17:17, 9 February 2013 (UTC)[reply]
Don't forget that, while with only a single frame or still picture, you can't enhance beyond what the pixels provide, in a video you can, to a limitted extent. If the subject, sized so that it/he/she covers only a few tens of pixels in any one frame, and it/he/she moves in a straight line a fraction of a pixel per frame (or a non-integer multiple of the pixel pitch), it can be enhanced. By means of computer software that moves it back at the same rate, in effect the number of pixels is increased, as edges in the subject cross from one pixel to another in a sort of picket fence effect. The improvement is not dramatic however - those movie simulated enhancements remain impossible. Two factors limit how good it is: 1) it cannot overcome optical blurring, and 2) the subject does not restrict their movements to convenient precise speed straight lines. Wickwack 58.167.247.139 (talk) 00:08, 8 February 2013 (UTC)[reply]
Thanks for the replies, everyone! And especially Nimur and StuRat! Evanh2008(talk|contribs) 14:30, 8 February 2013 (UTC)[reply]
Resolved
Hehe, putting that tag on an answer is like asking "what could possibly go wrong?" Here you go: [1]Wnt (talk) 16:52, 11 February 2013 (UTC)[reply]
biology
Assume there exists a 13 year old girl, capable of reproduction but still in the process of developing secondary sex characteristics, and she got pregnant. Will the girl continue developing? When will she continue developing? If she still lacks underarm hair, will her body produce such hair even if she got pregnant? — Preceding unsigned comment added by 49.145.4.37 (talk) 09:27, 7 February 2013 (UTC)[reply]
Have you seen any 40 or 50 year old women that appear to have stopped developing secondary sex characteristics at the age of 13? --Guy Macon (talk) 18:39, 7 February 2013 (UTC)[reply]
How is your posting even remotely responsive to the question? Or was it just intended as humor?75.34.25.6 (talk) 19:54, 7 February 2013 (UTC)[reply]
The post did answer the question. The more direct answer is, pregnancy does not halt puberty. thx1138 (talk) 20:27, 7 February 2013 (UTC)[reply]
Think. The end of puberty is commonly considered to be reproductive capability and achievement of maximal height. It is not "final attainment of secondary sex charateristics" because things like androgenic hair changes continue throughout life. Breasts increase size and change shape with both pregnancy and obesity. So yes, it is possible for a girl to be ovulating while she still has an inch of growth left, but both of those things are dependent on estrogen and are unlikely to be discordant. Pregnancy results in a further jump in estrogen levels and would bring growth to a close faster. alteripse (talk) 23:50, 7 February 2013 (UTC)[reply]
The above opinions are interesting. But do we have any biologers or farmists here who might be able to refer to scientiary sources that actually discuss whether pregnancy causes growth or stunting in immature animals? μηδείς (talk) 11:36 pm, 7 February 2013, last Thursday (3 days ago) (UTC−5)
The question was about girls, not animals, so I gave an answer about girls, not animals. Puberty is different enough in humans that even primates are not necessarily going to help with this, let alone farm animals. The following facts are not "opinions" and are actual data gathered on American girls about 50 yrs ago. Average bone age at menarche: 12y6m Average remaining growth at menarche: 2" if no pregnancy occurs. Many, not all, girls will ovulate within the first months after menarche. Some even do it once before menarche. Here are some citations on what happens to growth when pregnancy occurs before growth is complete. http://www.ncbi.nlm.nih.gov/pubmed/8262493http://ajcn.nutrition.org/content/60/2/183.long Look for a farmer or biologist if you like, but there is no type of person more professionally qualified to answer this question than a pediatric endocrinologist. alteripse (talk) 22:23, 8 February 2013 (UTC)[reply]
Interesting source, but the specific fact I was asking about was the difference in final height between girls who do and do not have a pregnancy while still growing. That paper was largely about fetal weight, and did not seem to compare the two populations--unless I missed it. You say 2" inches is expected after menarche with no pregnancy--doo you have the figure for how many inches are expected after menarche if there is a pregnancy? μηδείς (talk) 01:47, 9 February 2013 (UTC)[reply]
No, because it will depend on how much remaining growth when she gets pregnant. I know of a pregnancy that occurred even without menarche (i.e, the first ovulation was fertilized) in an 11 yr old mildly precocious girl who had about 3" of remaining growth by bone age prediction at the beginning of the pregnancy. The recent reports of the 9 yr old Mexican girl and of course Lina Medina are examples of more than 2" of remaining growth potential at the time of pregnancy, but i have seen no data for them on height at impregnation and height eventually attained. For obvious reasons, pregnancies in girls with significant amounts of remaining growth are rare and usually terminated, so it is typically difficult to find good data. The closer to average timing the girl's puberty is, the more likely that bone age and menarche will be concordant, and the likely height loss less than an inch. It is in a girl whose puberty is precocious that there may be more chance of losing height growth, but these are less likely to carry to delivery, and the average doctor's office height measurement is only accurate to the nearest inch in many cases, making it difficult to detect changes of growth potential unless by chance there had already been an endocrine evaluation. The camden paper discussed nutritional limitation of fetal growth and confirmed that at least some height growth continued in the girls, though measured stature was compromised by the lordosis of pregnancy so they were assessing growth by knemometry. They did not try to assess the difference between bone-age-based height prediction at diagnosis of pregnancy and eventual attained height well after delivery, which is not exact but would be the best way to answer your question. The reason we can assume that there would be at least a little height loss from estrogenic acceleration of bone maturation is that pregnancy estrogen levels are comparable to those used to accelerate bone maturation and attenuate adult height in girls thought to be too tall or in the Ashley treatment. The more common situation in which this issue comes up is when oral contraceptives are being considered in a girl who has not finished growing; potential height loss is one of the factors weighed. alteripse (talk) 02:15, 9 February 2013 (UTC)[reply]
Thanks for the detailed answer, I had assumed some stunting would be the likely case, but figured paradoxically there might be a growth spurt. The effect of oral contraceptives makes sense. μηδείς (talk) 02:26, 9 February 2013 (UTC)[reply]
Unseen effects on water when going down a drain
What causes water to swirl when going down a drain? Does it rotate in one direction in the northern hemisphere and the other direction in the southern? Question from Sir Lar Feb. 7,2013 Payson Az. 85541 — Preceding unsigned comment added by Sir Lar (talk • contribs) 10:29, 7 February 2013 (UTC)[reply]
There's a nice discussion of this here: Coriolis effect#Draining in bathtubs and toilets. Basically, unless conditions are very carefully controlled, the effect of any initial movement of the water or irregularities in the container will predominate over the Coriolis effect, which is what would cause the rotation to be the opposite direction in each hemisphere. Equisetum (talk | contributions) 11:44, 7 February 2013 (UTC)[reply]
The business of clockwise versus anticlockwise sink/bathtub swirl in the northern and southern hemispheres is a complete and utter myth. It's quite amazing to me that so many people believe this when the most trivial of experiment will demonstrate that it's clearly not true - just watch your bathtub and sinks for a few days - you won't see any consistency whatever in which way they swirl. The magnitude of the coriolis effect is vastly too small to produce this effect. Even the slightest motion of the water or irregularity in the shape of the container or the drain will easily overwhelm it. Coriolis only has an effect for things that move LONG distances in the north/south direction - things on the scale of ocean currents, hurricanes.
The interesting part of this question (and I don't immediately see an answer) is why water swirls as it goes down the drain at all.
had one of those deeply mystical latenight college discussions on related topics once. our final decision was that spirals and/or helices are the normal state, straight lines and circles are the singular variations on the general theme. Consider; with repeating subunits, whether physical or force, there is always an angle between the new vector and the previous, which can be projected onto a vector parallel with the original, and a vector in a plane perpendicular; i.e. a forward motion, and a rotation. if the former is 0, it's a circle. if the latter is zero, it's a straight line. otherwise, a helix. thus the helical config of proteins, DNA, etc., is just an intrinisic feature of repeating subunits, not anything specificall evolved. in this case, you could sort of imagine a centripetal force, as in this case due to the "perpendicular plane" being a bowl, sort of reminiscent of Einstein's rubber sheet space topology, it's a spiral in/down, really a shallow helix/spiral combo. Gzuckier (talk) 15:09, 7 February 2013 (UTC)[reply]
I'd say spiral water flow down a drain is a result of both an unbalanced downward force and an unbalanced inward force. The downward force is easy enough to visualize, as the open drain leaves an unsupported column of water above it, which then falls down the drain. This then leaves a gap in the water, and water pressure then pushes the water in from the sides, providing the inward force. In reality, they don't happen in distinct steps like that, and the water is moving downward as it moves inward, so you never quite replace all the water, at least if the distance from the drain to the surface is small, leaving an indentation in the surface of the water, possibly extending all the way down to the drain. As for which way it chooses to rotate, I agree that any minor asymmetry can start it going one way or the other. StuRat (talk) 16:25, 7 February 2013 (UTC)[reply]
I kinda suspect some kind of conservation of angular momentum thing going on here. The huge body of water has a certain angular momentum just due to random motion in the container - but as it heads down the drain, the flow is straightened out. The conservation law might therefore imply that the angular momentum ends up being concentrated in a smaller and smaller mass of water - so the rotation rate has to go faster and faster as the container empties. That's a guess - but it might explain why the swirl goes faster and faster as the water flows away. SteveBaker (talk) 17:25, 7 February 2013 (UTC)[reply]
No need to speculate: this is a standard homework problem in any fluid-mechanics class. Here's a nice website from MIT: Conservation of Angular Momentum (in the context of an aerospace fluid mechanics class). If you refactor this as part of a flow equation, you can easily see that angular momentum effects an equivalent pressure that "pushes" the water away from the center of a vortex. This is analogous to effective gravitational potential. You can refactor the law of conservation of momentum so that in the equation, it looks like it's creating a net force, or a net pressure, or effectively creating an energy barrier, to keep the water from flowing in to the center of the vortex, in contraposition to gravity (which wants to make the water fall into a lower position, in the vortex). (This pressure is the ensemble effect of the "fictitious" centrifugal force on each fluid molecule). That's how you can sustain a "vertical edge" to the water. Nimur (talk) 01:14, 8 February 2013 (UTC)[reply]
Yes, any large body of gas or liquid will have some small random net spin (the chance of it not is astronomical) and hence, due to conservation of angular momentum spin faster as it contracts. μηδείς (talk) 19:02, 7 February 2013 (UTC)[reply]
It's not really necessary to go beyond the mere fact that any mass of randomly moving particles is going to have some sort of net spin. The only way it wouldn't would be if every particle's motion miraculously happened to cancel out. Either there will be a small amount of particles on the right moving faster toward the drain, or on the left, and voila, a vortex starts spinning. μηδείς (talk) 01:35, 8 February 2013 (UTC)[reply]
Right - so (naively) if (say) 60 liters of water in an idealized 2 meter diameter cylindrical bathtub is rotating at a leisurely one revolution per minute (almost too slow to notice) then when it's drained down to the last liter, it'll be spinning at one revolution per second - and as the last 1/10th liter disappears, ten revolutions per second. But real-world bathtubs have a smaller cross-sectional area at the bottom - so when the water is spinning around in a 10cm circle centered on the drain-hole, it'll be zipping around at 100 revolutions per second! Obviously this is a very naive estimation because friction, viscosity and the irregular shape of a typical bathtub would change all of that - and also, the entire body of water isn't rotating at the same speed and some residual angular momentum surely resides in the water as it heads down the drain. But you can easily see how that rotation can go from negligable to dizzying as the water drains out. This also explains why the speed of the swirl is so much lower in a handbasin than in a bathtub.
Hmmmm....now I want to watch a swimming pool drain out! SteveBaker (talk) 17:39, 8 February 2013 (UTC)[reply]
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Reventazón River vs. Reventado River in Costa Rica
First of all, feel free to move this to the Miscellaneous desk if geography isn't science-y enough. So, my question. There's the Reventazón River in Costa Rica. This river is not covered in Library of Congress Subject Headings, but LCSH does have a "Reventado River (Cartago, Costa Rica)." I suspect these both refer to the same river. I looked up Lake Cachí in Google Maps, and it's within Cartago Province. So it doesn't seem likely that two rivers with such similar names would be in the same province. Our List of rivers of Costa Rica doesn't mention a Reventado, nor does its equivalent in Spanish Wikipedia. I suspect the LCSH term is an older name for the river, but I'm trying to confirm this. I'm waiting to hear back from the Library of Congress about this, but I'm impatient. --BDD (talk) 21:59, 7 February 2013 (UTC)[reply]
Michigan has eight Pine Rivers and seven Black Rivers. Name similarity is a poor standard. Rmhermen (talk) 23:13, 7 February 2013 (UTC)[reply]
For posterity's sake, I heard back from the Library of Congress, and these are indeed different rivers. --BDD (talk) 17:14, 11 February 2013 (UTC)[reply]
February 8
Camera quality as a function of weight
When taking a photo using a DSLR camera , is there a difference between a heavy camera and a light camera if the photo is taken handheld and all other factors being the same? — Preceding unsigned comment added by 149.135.147.66 (talk) 02:02, 8 February 2013 (UTC)[reply]
I'm no expert, but in film SLRs, heavier is generally better because of all of the film mechanism needing to get it in the correct place, etc. In a DSLR a heavier camera might be better to damp out some of the jiggling when it is hand held. Bubba73You talkin' to me? 04:01, 8 February 2013 (UTC)[reply]
Maybe in a certain range, but there's also a point where things become so heavy we lose the ability to hold them steady. StuRat (talk) 04:23, 8 February 2013 (UTC)[reply]
But the OP specifically said handheld which I would take to mean without using a tripod or monopod or other supporting device. They also said 'all other factors being the same' which is a big unclear but I take to mean that they're excluding the heavier camera possibly having a better lens, CCD and whatever else. Nil Einne (talk) 06:02, 8 February 2013 (UTC)[reply]
Plenty of sports photographers use monopods on their hand-helds, to help them steady the camera, especially on zoom-in shots. ←Baseball BugsWhat's up, Doc?carrots→ 14:00, 8 February 2013 (UTC)[reply]
But how much does it help when the problem is the camera is too heavy? Nil Einne (talk) 14:42, 8 February 2013 (UTC)[reply]
The monopod (and hence the surface it's resting on) absorbs much of the weight of the camera, hence it's easier to keep steady. ←Baseball BugsWhat's up, Doc?carrots→ 22:20, 8 February 2013 (UTC)[reply]
The pictures that I take my fixed lens compact film camera, are much crisper than those I take with the compact digital. With the film camera, I look through the view finder and the camera sits against my face, which helps to hold it still. With the digital, I have to hold it at arms length to see the viewing screen and it is very difficult to hold the camera still. The pictures that I take with the film SLRs are the crispest of all, as the cameras have size and you can get a proper grip. With small cameras, I have to hold them in my finger tips, otherwise part of the lens, or sensor, or view finder is obscured. I don't have a DSLR but I imagine the same it true. --TrogWoolley (talk) 09:30, 8 February 2013 (UTC)[reply]
The mechanics of holding cameras is probably part of the resolution difference, but compact digital cameras have poor resolution. Bubba73You talkin' to me? 02:48, 10 February 2013 (UTC)[reply]
Anti-cancer molecule TRAIL
TRAIL was recently in the news concerning a new approach to fight cancer. I have a few questions about it:
Which cells produce TRAIL, and under what circumstances?
Does TRAIL act on the receptors of the cell that produced it, or on some other cell?
Is TRAIL a transmembrane protein, or is it released into the extracellular fluid? (Our article states that it has "characteristics of a transmembrane protein".)
If the answer to any of these questions is "not currently known", that would be helpful too. Thanks, AxelBoldt (talk) 04:11, 8 February 2013 (UTC)[reply]
The NCBI is a reliable source and has a page about TRAIL. TRAIL is expressed in most tissues, in various forms, but appears to cause apoptosis of transformed and tumor cells more than others. As a cytokine, it is secreted. Like TNF, it causes inflammation. -- Scray (talk) 04:23, 8 February 2013 (UTC)[reply]
MTBE
Now that MTBE has been mostly phased out as a gasoline additive, are there any large-scale uses for it left? 24.23.196.85 (talk) 06:38, 8 February 2013 (UTC)[reply]
As an additive to petrol or gasoline? That's what the article says.... So does this press release [2]Nil Einne (talk) 06:52, 8 February 2013 (UTC)[reply]
It is. That's why I said 'or' rather then 'and'. MTBE is evidently still used to some extent in a few countries like India [3], Singapore [4] and Malaysia [5] where the word petrol would be preferred. It's also used in some countries where the word gasoline might be preferred (well I'm not sure of this). I originally just used the word petrol, but decided to add gasoline to avoid confusion and argument. Nil Einne (talk) 07:41, 8 February 2013 (UTC)[reply]
MTBE is still used in Mali. There is a phase-out plan[6] but the old refineries are still running. Let me pull up my cheatsheet just in case anyone here is thinking of taking a vacation in Timbuktu...
Fuel of any type: Carburant.
Gasoline/petrol: L'essence or Essence
A sentence with Sélectionnez and carburant means "choose your fuel" Usually there is an octane number but you may see ordinaire or super.
A sentence with Validez and carburant means "confirm" (press "Val" to confirm.)
Unleaded: Essence sans plomb
Leaded: Essence au plomb
Diesel: Gazole or Gasoil
Liquid petroleum gas (LPG): GPL or Gépel
Heating oil: Fioul or sometimes Fuel
Crude Oil (and, I think, paraffin): pétrole
The last two come into play when you ask for petrol or fuel; they will say that they don't have any. Ask for gasoline and you may get essence or you may get gasoil.
Are we talking about the same thing? The link you supplied is referring to Tetraethyllead which MTBE partially replaced. If they still haven't phased out Tetraethyllead in Mali, I'm not sure whether there are any concrete plans for phasing out MTBE (although the link you supplies is rather old). In any case, MTBE is still widely used in China. As I mentioned it is apparently used in India, Malaysia and even I think Singapore (I'm not 100% sure since the results get confused by level of discussion of production and blending in Singapore but Singapore is apparently a major importer). These are places (including China) which phased out tetraethyllead at least 13 years ago. It's also used in quite a few other Asian (including Arab) countries. I'm not sure if there is a clear phase out plan in any of these but as per the sources its use is apparently decreasing in some cases although because of price pressures not environmental or health concerns (apparently in China this includes a consumption tax [7]). However the sources suggest while production may not be increasing in worldwide terms (or at least not that much), it's not decreasing either. See also [8]. Either way all the evidence suggests MTBE is still widely used as a gasoline/petrol/whatever additive. P.S. Our article suggest TEL was phased out in Africa in 2006 although Algeria remains a holdout and there are a bunch of Asian countries and one arguably European country (Georgia) who are still using TEL. I suspect there are more who have some leaded petrol but don't really know. Nil Einne (talk) 12:36, 8 February 2013 (UTC)[reply]
You are right. I confused the two. (Note to self: next time, smoke crack after editing Wikipedia...) --Guy Macon (talk) 19:32, 8 February 2013 (UTC)[reply]
Wait a minute, how can Georgia be an "arguably European country" when in fact it's part of the USA? 24.23.196.85 (talk) 22:57, 8 February 2013 (UTC)[reply]
Nil Einne was referring to საქართველო (that's Sakartvelo for you illiterates). -- Jack of Oz[Talk] 01:04, 10 February 2013 (UTC)[reply]
Oh, that Georgia! In that case, they might well have different fuel specs from the ones in our country. So, getting back to the topic of MTBE, I gather that its only major use is as a fuel additive? 24.23.196.85 (talk) 04:38, 10 February 2013 (UTC)[reply]
Long time later but in retrospect petrol/gasoline would have been clearer. Nil Einne (talk) 21:57, 20 July 2013 (UTC)[reply]
Jupiter and the Romans
The ancient Romans gave Jupiter its name after their main god. Did they know that it was the biggest planet in the solar system, or is it just a coincidence? It's certainly not the brightest object, or even the brightests planet visible in the northern skies, so it's not immediately clear to me why they assigned it such importance unless they knew it was so large. 202.155.85.18 (talk) 07:47, 8 February 2013 (UTC)[reply]
It is usually the brightest planet in the night sky. (Venus is seen only before sunrise or after sunset.) Ruslik_Zero 09:08, 8 February 2013 (UTC)[reply]
I was taught at school that they believed that these objects in the heavens (asters) actually were gods, like the Sun, which is the most obvious godly thing. Many certainly must have thought that Jupiter-the-God and Jupiter-the-aster were the same thing or that one was a representation of the other, but wikipedia does not explain that relation very well, for example our article Jupiter (mythology) doesn't mention the words astronomy or astrology at all. As to why it became so important in the mythology is not related to its actual size (which they could not calculate very accurately), but to the stories in the myth. --Lgriot (talk) 09:21, 8 February 2013 (UTC)[reply]
No, there's no way they could have known. No planet is big enough to show a disk to the naked eye, so they couldn't have measured any planet's angular diameter.* They also couldn't have measured distance because the only way was by parallax, but even the parallax of Mars at closest approach is less than 1 arcminute from opposite sides of the Earth.
Not only is Jupiter usually the brightest planet, it's also brighter than every star. Venus is not always visible, even in the morning or evening; it's often hidden in the glare of the Sun, although the Romans would have known this instead of thinking it disappeared.
*Actually, there is one easy way to measure angular diameter. Hold your head absolutely still, and watch as a planet disappears behind a building due to Earth's rotation. A star would disappear instantly because its angular diameter is tiny, whereas planets take a few seconds. The longer it takes, the larger the planet's angular diameter. If you're patient enough, you can also use the Moon instead of a building (see lunar occultation). --140.180.247.198 (talk) 10:37, 8 February 2013 (UTC)[reply]
That will fail for a number of reasons, including atmospheric blurring, but most importantly the fact that it is nowhere near possible to hold your head still enough. Looie496 (talk) 16:49, 8 February 2013 (UTC)[reply]
You've obviously not seen A Clockwork Orange. μηδείς (talk) 17:33, 8 February 2013 (UTC)[reply]
Interesting... I feel like there must be a way. In theory, if you have a thin piece of wood with slits of gradually increasing width through it, you should be able to move your head, looking at the images through the piece of wood, until they stop getting brighter. I have no idea if the ancients did this (or, more likely, some much more clever version) Wnt (talk) 19:04, 8 February 2013 (UTC)[reply]
I don't understand the Clockwork Orange reference, but atmospheric blurring is nowhere near enough to make a star look like a planet. On a normal night, astronomical seeing produces a disk around 1 arcsecond across, whereas planets are tens of arcseconds across. As for holding your head still, you can use a sufficiently distant object--a few hundred meters is more than enough if you can hold still to within 1 cm--or you can use a vice to clamp your head. --140.180.247.198 (talk) 19:37, 8 February 2013 (UTC)[reply]
The "vice to the head" thing was the Clockwork Orange reference. The protagonist's head being held in a vice-like apparatus and being forced to watch images is one of the iconic scenes from the movie (if not from the book). See the image at A_Clockwork_Orange_(film)#Psychology. -- 205.175.124.30 (talk) 19:48, 8 February 2013 (UTC)[reply]
than those of lighter isotopes?--Inspector (talk) 12:23, 8 February 2013 (UTC)[reply]
What do you mean by "heavier elements"? The answer is yes for that part, if I understand you correctly. By volume, they are heavier in their elemental state. The compounds would be heavier, unless they combine with lighter isotopes of other elements. Otherwise, yes, the compounds would be heavier, since (as I understand it) the properties of isotopes are more or less the same, other than molar mass and stability. IBE (talk) 13:33, 8 February 2013 (UTC)[reply]
We have a whole article about isotopes, covering the differences and similarities. DMacks (talk) 16:36, 8 February 2013 (UTC)[reply]
Check whether the following two points are right or wrong.
1. Sound energy is a form of energy, but sound is a mechanical wave (not energy). Light is neither energy nor a form of energy, but it is an electromagnetic radiation. Radiant energy is the energy of electromagnetic waves. Heat is not a form of energy, but heat itself is energy.
2. Energy of waves or radiations is directly proportional to frequency and inversely proportional to wavelength.
I don't know whatever I have written is right or wrong. Kindly correct the wrong statements. Sunny Singh (DAV) (talk) 12:43, 8 February 2013 (UTC)[reply]
Sound, light and heat are all forms of energy. You could probably do a bit of reading first, and post a couple of references to articles which didn't quite answer your questions. IBE (talk) 13:35, 8 February 2013 (UTC)[reply]
Not sure about this usage of "forms of energy". Energy is an attribute of certain physical phenomena and systems. Saying "sound, light and heat are all forms of energy" is like saying "my blue shirt, my red tie and my brown shoes are all forms of colour". Gandalf61 (talk) 13:47, 8 February 2013 (UTC)[reply]
The easiest way to think of energy is "the ability to cause a change". The important things about this ability is that it is a) quantifiable and b) conserved. That means that you can measure it, and that it doesn't go away, nor is it created (though it can become unusable, see entropy and second law of thermodynamics). The various "forms of energy" are the (somewhat arbitrary) way in which we organize or categorize the various kinds of changes. Like all categorization schemes, these aren't rigidly defined categories which are inherent in the system itself, but are human-created distinctions which allow us to extract more understanding about the system. So, we can talk about energy in terms of "kinetic" and "potential" forms (that is, broadly speaking, energy which is currently causing a change, and energy which is "stored up", waiting to cause a change at a future time); or we can talk about it in terms of the types of changes that occur (mechanical energy, chemical energy, light energy, heat energy, etc.) None of these definitions have bright line boundaries; they're all a little fuzzy around the edge, depending on the specific model you are using. It's important also, to note that energy is not a kind of stuff you can hold in a bottle. It's a way to quantify changes. The way in which energy, time, and matter work together is the basic definition of the science of physics: physics describes how every kind of change occurs through mathematical models. Wave energy is just energy which occurs in a repetitive pattern. Some waves are obvious (sound waves, ocean waves). There are some phenomena which can be modeled as either wave energy (oscillations of a field rather than oscillations of collections of particles) or as particles-in-motion; the fact that these phenomena could be modeled both ways is called wave-particle duality. Energy is a really complex concept, so I hope this sort of explanation is helpful. The iconoclastic physicist Richard Feynman, easily one of the most intelligent and articulate physicists of the last hundred years readily admitted that even he couldn't adequately understand nor define what energy really is. I hope that helps some. --Jayron32 14:26, 8 February 2013 (UTC)[reply]
We end up always defining something by its properties. Defining something by what it does (like energy is the "the ability to cause a change") seems to me to be the standard way of defining things. You define 'brown' by the wavelength that is reflected by a brown object. "Function" is "something that takes one value and outputs another. Knife is something that cuts, and can be hand-held. OsmanRF34 (talk) 15:47, 8 February 2013 (UTC)[reply]
What's the difference between "form of energy" and "energy"? If it's energy (like sound, light), it has to be in some form. If it's not a form of energy (like length) it cannot be energy. OsmanRF34 (talk) 14:23, 8 February 2013 (UTC)[reply]
Article heat starts with this "heat is energy transferred from one body to another by thermal interactions". This made me think that heat is energy but not a form of energy. Jayron you wrote heat energy, but this answer says writing heat energy is bad english. This is where the confusion starts. Make it clear. Sunny Singh (DAV) (talk) 17:02, 8 February 2013 (UTC)[reply]
Meh. Heat energy, thermal energy. Whatever. The language is a bit fuzzy in this regard; it depends entirely on which text book you're reading that day. Also, your sentence " heat is energy but not a form of energy" is actually logically silly: a form of something is merely an example or category of it. You can't simultaneously be something, and then not be the same thing. Heat (or thermal energy, if you prefer) is merely the energy of molecular motion. Some texts will draw the distinction between heat and thermal energy insofar as heat is the transfer or movement of that thermal energy. In that way, heat is to thermal energy as work is to mechanical energy. One very common expression of energy is the equation ΔE = q + w; that is ΔE (the total energy changes or transfers in a system) is equal to the sum of q (the heat) and w (the work). So heat just means "how thermal energy moves around" in the same way that work means "how objects move around"; and one way to think of energy changes in systems is to think of such changes as either heat or work or some combination thereof. --Jayron32 17:37, 8 February 2013 (UTC)[reply]
So, this means saying heat or heat energy; light or light energy; sound or sound energy, all are same. Sunny Singh (DAV) (talk) 05:14, 9 February 2013 (UTC)[reply]
.
Upon reading Sunny Singh's question here and last time (Nov 14 2012), and the answers, I think that folk have not percieved Sunny Singh's fundamental problem, though this time people have come close to one part of it. Sunny Singh has two areas of difficulty: a) he's not too good on the English language, and b) he doesn't have a clear idea of what energy is. To a certain extent we have to guess what Sunny wants of us, and what he needs of us, because his English is somewhat cryptic. It would be helpfull if Sunny responds to this posting and lets us know if I and the other posters are on track or off track.
First, some points on English: It is indeed poor English to say "heat energy", "light energy" and "sound energy", though you will find those phrases in lay writing and occaisonally even in professional text books. The reason why it is poor English is the same as it is poor English to write "cat animal", "goat animal", "snake animal" - becaue they ARE all animals - each phrase is much like writing "animal animal". The only difference is, everybody knows cats, goats, and snakes are all forms of animals, but those of us without a good science education may not know that heat, light, and sound are all forms of energy. Got the idea? Just as we write "cat" when we mean cat, and not "cat animal", we write "light" when we mean light.
Before answering the specific questions you posted, Sunny Singh, let's try and make what energy IS, very clear. OsmanRF gave you a key. Energy is that which can do something or change something. Energy is something that can be accumulated or converted - energy mathematically is the product of power and time (or if you like, intensity x time). For example: an electric lamp rated at 50 watts, when operated for 10 seconds, consumes 500 joules of electricity (another form of energy) and converts it into light and heat, both emitted to a total extent of 500 joules. Another example: A loudspeaker is fed from a stereo amplifier playing a pure tone at 10 watts for 1 minute. That loudspeaker has consumed 600 joules of electricity (from the amplifer) and converted it into a total of 600 joules of heat and sound. If the loudspeaker was 100% efficient, it would emit 600 joules of sound, and no heat. Get the idea? Energy is somthing that, over time, gets converted from one form to another.
Now, we can look at your posted questions, Sunny. As I said, your English is cryptic, so I've altered them a bit here and there to mean what I think you meant to say. Let us know if I guessed wrong on what you meant.
1a Sound is a form of energy - correct.
1b Sound is a mechanical wave, and a mechanical wave is not energy - This statement is wrong because a mechanical wave is a form of energy, because a transducer, such as a microphone, can convert it over time into another form (electricity).
1c Light is not energy - this statement is obviously wrong
1d Light is electromagnetic radiation - Yes it is, and electromagnetic radiation is another form of energy - by suitable means light can be converted over time into another form of energy - e.g., into heat by means of a black object.
1e Heat is not energy but heat itself is energy - This statement is wrong because heat is a form of energy - it can be converted into another form of energy (light, sound, etc).
2. The energy in waves or radiation is directly proportional to frequency and inversly proportional to wavelength. - This statement is completely wrong. I'm guessing Sunny included it because he was confused by reading physics texts while not having a clear understanding of energy first. For any sort of wave or radiation (i.e., mechanical waves such as sound, or electromagnetic radiation such as radio waves), the wave has magnitude, or intensity, measured in watts. Remember, energy is magnitude multiplied by time. Frequency and wavelength has NOTHING to do with it. If 2 watts of sound is coming out of a loudspeaker, then that loudspeaker is emitting 2 joules per second of energy, regardless of frequency. Confusion for lay people can come when discussing electromagnetic energy, which is a combination of electric and magnetic fields. Physicists don't realy know what electric and magnetic fields are. But we have long learnt that for many purposes the mathematics of waves gives the right answers, and for other purposes the mathematics of particles (photons) gives the right answers. As the theory of electromagnetic radiation is a dual of waves and photons, we can combine the two and say things like "photons are emitted at a frequency of x hertz". The higher the frequency the more energy in each photon. Note that saying "photons are emitted at 20 megahetz" (1 megahertz means 1 milion cycles per second) DOES NOT mean 20 million photons emitted per second. The higher the frequency, the fewer photons needed. The greater the power, the more photons are needed. Each photon is a packet of energy. The amount of energy in each packet is proportional to the frequency of the equivalent wave.
Energy is an attribute of physical phenomena and systems. A thrown ball has various attributes, including height/altitude, velocity, kinetic energy and potential energy. We can say "the ball has energy" but it is incorrect to say "the ball is energy" or "the ball is a form of energy". This would be like saying "the ball is velocity". Similarly, a sound wave or an electromagnetic wave has a frequency, a wavelength, an amplitude and an energy level (or, more generally, it has a spectrum of these attributes). We can say "a wave has energy" but it is incorrect to say "a wave 'is energy" and also incorrect to say "a wave is a form of energy". Gandalf61 (talk) 11:57, 9 February 2013 (UTC)[reply]
No, that isn't right. You are correct in saying that it is wrong to say a ball is energy. It is clumsy to say that a ball of mass m at some height z is a packet of potential energy. But the analogy does not stretch to waves. A fundamental propery of waves/radiation is magnitude/intensity. The energy of a wave is merely the product of magnitude and time as I said. Any two and you can calculate the third. Thus the three things, magnitude, time, and energy, are not independent attributes. Wavelength can be calculated from frequency and velocity of propagation (which is determined by the medium) - so these are not independent of each other, either. However, frequency/wavelength/velocity IS independent of magnitude, energy and time. When a wave passes from one lossless medium to another, neither magnitude nor frequency is changed. Thus, if we want to boil waves/radiation down to the simplest set of fundamentals, there are two fundamentals: magnitude and frequency (not counting direction of propagation and polarisation). These two attributes completely describe the wave. These two attributes, along with identifying what sort of radiation it is, completely describe the radiation. So, it is perfectly ok to use the energy describing words "light", "sound" etc to mean energy in the respective energy form. There is no need nor sense in saying things like "the energy in the wave." One does not say "the radio transmission has an energy of 200 joules per second" (this is much the same as why it is bad English to say "sound energy", "light energy" etc); we can just say "the radio transmission IS 200 joules per second." (More normally we would say "a 200 Watt transmission"). It is entirely correct to say "sound is a form of energy", "a wave is a form of energy", etc. Wickwack 60.230.225.90 (talk) 13:23, 9 February 2013 (UTC)[reply]
Most of that is nonsense. Obviously a wave is not completely described by its magnitude and frequency alone. A wave has many other attributes, some of which you mention and then randomly discard. And you seem to be forgetting that most waves do not even have a single frequency. No-one says "the radio transmission IS 200 joules per second" or "the radio transmission is 200 Watts". The phrase "a 200 Watt transmission" is obviously short for "a transmission that has a power of 200 Watts". Saying "sound is a form of energy" makes as little sense as saying "my shirt is a form of colour" or "a car is a form of momentum". Sound energy is a form of energy, but the sound wave itself is neither energy nor a form of energy - see our article on forms of energy. Gandalf61 (talk) 15:57, 9 February 2013 (UTC)[reply]
I just love people who provide links to Wiki articles or web pages, but those articles do not support their argument - as is the case with Gandalf's post. You have not read or understood the word "fundamental" in my post above. As I said, the two fundametal properties of waves/radiation are magnitude and frequency. All the other parameters you mentioned, wavelength, amplitude, energy, and more, can be calculated from the two fundamental properties plus time and velocity. Velocity is not a property of the radiation, it is fixed by the medium. I haven't discarded anything. For example, let's say a source is emitting in vacuum a light bean of magnitude 200 W at a frequency of 6 x 1014 Hertz. The energy is 200 joules per sec. The velocity is the speed in vacuo, ~ 300 x 106 m/sec. Thus the wavelength is 300 nanometre. Now lets say the beam passes from vacuum into a block of perfect silica glass. In this glass the velocity is only about 150 x 106 m/sec (I've tweeaked the value a bit to make the meantal arithmetic easy). Magnitude is still 200 Watt. Frequency is still 6 x 1014 Hertz. Energy is unchanged. Wavelength is doubled. See how it works? See that magnitude and frequency are fundamental to the wave, and all the other attributes can be calculated? Yes, more than one frequency may be present. That does not invalidate what I said - it merely means there is more than one wave superimposed in the medium or space. The total amplitude can be calculated from the addition of the component waves. Yep, nobody says "the radio transmission is 200 joules per second", usually. It is perfectly valid to do so though, and sometimes, when the energy is of primary interest, we do just that. It is just more convenient in most situations to say "a 200 Watt transmission.
Now, what is different about saying "a car is a form of momentum", and "sound is a form of energy"? Lots. Momentum is is just one possible attribute of a car - a car has a vast mumber of independent attributes - height, number of doors, colour of paint, etc etc. A particular car may have at some point in time, a lot of momentum, a litle, or no momentum at all. Regardless, it's still a car. But sound is completely described with a couple of fundamentals. If the sound has zero energy, there is no sound.
"Magnitude and frequency are fundamental to the wave, and all the other attributes can be calculated" ... more nonsense. For example, the phase of a wave cannot be calculated from its magnitude and frequency. Gandalf61 (talk) 05:45, 10 February 2013 (UTC)[reply]
True, sort of, but phase is not relevant to the question the OP asked. Phase is not an intrinsic property of a wave. You can only measure phase with respect to an independent timing reference. Keit 121.215.141.120 (talk) 05:56, 10 February 2013 (UTC)[reply]
Correct, Gandalf has not got a clear picture. If a light beam or radio beam in deep space passes a little green alien, he can say (translated into English and SI units) "I sense a light wave, frequency 6 x 1014 Hertz, magnitude 200 Watts." But he can assign to it any arbitary phase he likes, because the wave does not itself have a frame of reference. Of course phase cannot be calculated from just magnitude and frequency, as phase is not a wave property - it is a function of both the wave AND the separately existing timing reference. Wickwack 121.221.41.198 (talk) 06:38, 10 February 2013 (UTC)[reply]
Total nonsense. Of course phase is a property of a wave, and is just as "intrinsic" as its magnitude, frequency, wavelength etc. You cannot calculate the superposition of two waves from knowing only their magnitudes and frequencies. You also need to know the phase of one relative to the other. So the waves cannot be completely described by their magnitudes and frequencies alone. Your selection of magnitude and frequency as the only "fundamental" properties of a wave is both arbitrary and unscientific. Your "frame of reference" argument is nonsense because it leads you to the ridiculous conclusion that energy is not a property of a thrown ball because its potential energy is measured relative to an arbitrary ground level and its velocity (and hence its kinetic energy) is measured relative to an arbitrary frame of reference. Your whole chain of reasoning that you are making up to justify your incorrect statement that "sound is a form of energy" is flawed at every step. Gandalf61 (talk) 09:10, 10 February 2013 (UTC)[reply]
You are being silly now. If two waves are identical in frequency, you can determine a phase difference - really this is just using one as the frame of reference to the other. You cannot, without a third reference say one wave has this phase and the other wave has that phase. If two waves differ in frequency, the phase difference is constantly changing over time, and you still can't assign a phase to either one. Phase is NOT a property intrinsic to a wave. Nowhere have I ever said that energy is not an attribute of a ball, thrown or otherwise. It is one of a multitude of attributes of a ball. Go back and read more carefully. Wickwack 120.145.68.197 (talk) 09:55, 10 February 2013 (UTC)[reply]
But the logical conclusion of your own arguments is that energy is not an "intrinsic" or "fundamental" property of the ball since it is measured relative to an arbitrary frame of reference. And the same conclusion can be applied to the potential and kinetic energy of any other system - you cannot allow those to be "intrinsic" or "fundamental" properties either. Or chemical energy, which is also only ever measured as a relative difference, not an absolute quantity. I agree these are all silly conclusions - but they are the conclusions that result from your own flawed reasoning. By repeating your nonsense about "fundamental" properties and "sound is a form of energy" you appear to be trolling, so I am done here. Gandalf61 (talk) 11:53, 10 February 2013 (UTC)[reply]
No, that does not follow from my arguments. You have not recognised the difference: energy IS a property of balls and waves (over time), and does not require an external frame of reference, whereas phase does require an external frame of reference. The convention of asigning zero chemical energy to common simple chemical forms (e.g., O2, N2, C, etc), and relating other substances back to them is a matter of sensible convenience and is somewhat arbitary. Wickwack 121.215.66.37 (talk) 14:19, 10 February 2013 (UTC)[reply]
Both Wickwack and Gandalf61 are missing the importance of understanding and communicating context. Analogous to forms of energy might be forms of liquids, i.e. water. Water is not always liquid, but it is a liquid when used in context. To say water is a form of liquid is to say water has the property of a liquid. If X is a liquid then water is an instance of X. Same with forms of energy. Sound, light and heat are each forms of energy. Each are distinguished of course by other properties but they nevertheless each have the obvious property of having energy. Contrary to Wickwack's Bad English nonsense, we sometimes do need to make it known which property we are talking about. For instance we don't say "light" equals mc squared, but we might say that radiant energy does. In nontechnical usage, heat means simply the quality of being hot which merely correlates with heat energy. Heat is energy, but only when it is understood that energy is what is being talked about, because context matters... and when its lacking we often use terms like "heat energy" (per these sources) to make it clear when its possible. --Modocc (talk) 13:49, 10 February 2013 (UTC)[reply]
Yes, we do sometimes need to make it plain, especially if the words are intended for lay people - I believe I said so myself. That doesn't make what I said nonsense though. I was trying to address what I think is the OP's difficulty, which seems to be not a clear idea of what energy is. Yes, you and I agree that Sound, light, and heat are forms of energy - soemthing crucial to the OP's question. For some reason Gandalf does not agree, and he's wrong. Wickwack 121.215.66.37 (talk) 14:12, 10 February 2013 (UTC)[reply]
Modocc - if you are equating "is a form of energy" with "has the property of having energy" then is it correct to day "a thrown ball is a form of energy", "gunpowder is a form of energy" or "this book is a form of energy" ? If not, then can you explain why not (if possible, without resorting to Wickwack's mystical and debunked notion of "fundamental properties") ? Gandalf61 (talk) 16:26, 10 February 2013 (UTC)[reply]
In the broadest possible terms, with current physics and AFAIK, there does not exist any physical object which does not have mass-energy. Indeed, the molecules of every object has mass-energy. Thus, each of your statements are actually true. In fact, I recently burned some moldy books in a bonfire. Now those books I had are nonexistent (in the present), but their energy changed form though due to the law of conservation of energy. -Modocc (talk) 17:59, 10 February 2013 (UTC)[reply]
Okay, so with your usage every physical entity - sound waves, radiation, chemical substances, material objects etc. - is a form of energy. That is a consistent usage (unlike Wickwack's nonsense) but it is different from the definition of forms of energy in our article. Gandalf61 (talk) 10:36, 11 February 2013 (UTC)[reply]
The article lists mass as a form of energy along with Einstein's mass-energy equivalenceE = mc2 which defines the energy content of matter. -Modocc (talk) 12:18, 11 February 2013 (UTC)[reply]
I must say that the above debate is very dissapointing. If the OP has long since given up I shouldn't be surprised. The talk about the E = mc2 mass -energy equivalence, while not wrong, is hardly relevant to the OP's difficulty - it is just a distraction. Gandalf61 is, at best, very confused, and at worst, just a troll. Wickwack says sound, light, waves etc are forms of energy. This is consistent with reality, and agrees with the Wiki article cited more than once by Gandalf61. However, for reasons unfathomable, Gandalf61 claimed (last sentence in his 2nd post and elsewhere) that they are not. In that he is talking nonsence. Yet, in his last post above, he himself used the term "forms of energy". Looking for a way to rationalise Gandalf61's posts, about the only think I can think of in his claims about phase etc versus Wickwack's fundamentals, is that Gandalf61 does not understand the meaning of the word "fundamental".
Giving Gandalf61 the benefit of the doubt, i.e., assuming he is not trolling, and because some folk do think it means something different (simplicity sometimes), here is the directionary meaning of "fundamental": basal, serving as foundation, primary, an essential. In scientific and engineering work fundamental means that from which everything else relevant can be inferred or calculated. So, Wickwack's assignment of frequency and magnitude as fundamental attributes is correct. As he said, all other attributes of waves (eg energy, period) can be calculated from these alone. In fact that is basic stuff familiar to any Engineer. It is no different to describing a circle. A circle can be defined entirely by just it's radius. A circle does have other attributes - diameter, circumference, area. But these can be calculated from the radius. For a circle, you can assign any one of radius, diameter, circumference, area as a fundamental - the only one that you need quote to define a circle. In the same way, all you need to define a wave is 2 things: frequency and magnitude, though you can just as validly use another two, e.g., period and magnitude. Standard physics textbooks, eg Physics for Scientists and Engineers, Giaconelli (in 3rd edition, see Chpater 15 Wave Motion, especially 15-1 thru 15-4), say just the same thing.
As Wickwack said, phase is not an attribute of a wave. Without an external reference, phase has no meaning - I would have thought that was obvious.
Gandalf61's repeated discussion of balls has no value. Quite clearly, frequency and magnitude completely specify a wave (if the type of wave is known of course, eg sound, light, radio, etc), and if you have finite values for frequency and magnitude, you have a wave, and you have energy being transported. If you don't, there is no wave, and no energy transport. Clearly, waves are thus a form of energy - sound is a form of energy, light is a form of energy, etc. Balls are quite different. While you can completely specify sound with just the two fundamentals of wave energy (frequency and magnitude), you cannot specify a ball only by quoting its energy. As Wickwack said, take away (all the kinetic and potential, thermal, etc) energy, and you still have a ball. Take away the energy from a wave, and there is no wave. Balls have lots of other attributes - shape, size, mass, structure, etc. That's why you can't just say a ball is a form of energy, unless there is special context. Outside that context, it isn't. A ball is an object having size, shape, mass, etc, and sometimes some energy. It is quite silly of Gandalf61 to try and consider balls and waves as equivalent, in the context of the OP's question.
Gandalf61 does understand that energy is a property of matter such as balls, although he was uncertain and mistaken about the semantics of "forms of energy". Wickwack did a fairly good job of critiquing the OP's statements as requested, so I hope the OP is satisfied with his answer, but Wickwack's didactic pedantry about "sound energy" being poor English, even though we have an article titled "Sound energy" which is listed in the "Forms of energy" article was distracting too. Sure take away the intensity of a sound's energy and there is no sound, but that's equivalent to taking away a piano's mass, in which case you have no piano nor music. -Modocc (talk) 19:59, 11 February 2013 (UTC)[reply]
That seems rather contrived. Take away the mass of anything that has mass, and in practice you haven't got it anymore. A better analogy to Gandalf's arguments is take away the musical sounds, say by playing it in a vacuum. You've still got a piano. We don't actually write, normally, "sound energy", "light energy", "radio wave energy", etc. We just write "sound", "light", "radio wave". But we can't just write "ball" and expect people to think of it as energy. We can't just write "piano" and expect people to assume it is being played. Re-using one of Wickwack's analogies, we don't write "cat animal" because we know all cats are animals. You can't have a cat that is not an animal. There's nothing in a cat that isn't animal (well, maybe the oddd bit of grass in its' tummy that it ate because it felt ill). You can't have a piano without mass. But we do write things like "the black cat", because cats come in different colours, and we wanted to say what colour that particular cat is. We say "upright piano", or "grand piano" if we want to state what kind we mean. There's no need to say "sound energy" because there is no sound without energy, and (mechanical) energy is all sound is.
A piano played in a vacuum is rather pointless, for you will still have music from the vibrant sounds transmitted from the strings to within its boards, for sound requires a medium. The energy of any sound, at its most basic level, is simply just a select fraction of the kinetic and potential energies of the particles of the medium. Moreover, mass is not just any arbitrary attribute either, its energy. Therefore taking away sound energy (or sound if you insist) is taking away mass anyway. As for the assertion that we automatically think of waves in terms of energy (which people clearly don't always do, for I'm quite fond of body-surfing them) and that we don't think of warm-blooded cats as energy, well... its probably because we don't usually need to and cats are a tad more complicated than a waveguide. -Modocc (talk) 05:10, 12 February 2013 (UTC)[reply]
Like Ratbone, I don't think this "mass is energy", presumably your E = mc2 line, helps the OP either. And mass isn't energy, it requires conversion into energy, by some process such as nuclear fission. Apart from that, thank you: yes, cats are more complicated - that was exactly my point. If you wanted to describe a cat to an alien from the next galaxy, using some dot points, you probably would not mention its' kinetic or potential energy. Just had a horrible thought though - with enough cats, which do have both mass and elasticity, you could have a wave passing through them. Agreed, many people do not think of waves as energy - but only because they are ignorant in science. Many people, including highly intelligent people, think that the universe and everything in it, substantially in the form that it is today, was created in 6 days, a few thousand years ago. That does not make it right. Those of us with at least a half baked education in science, and those of us who are engineers, should know that energy, that is, an interchange of some kinetic and potential (or elastic) energy as you said yourself (for mechanical waves) or the equivalent in electromagnetic waves, is what waves are. Wickwack 124.182.165.35 (talk) 07:25, 12 February 2013 (UTC)[reply]
Wickwack - I love the bit where you say "mass isn't energy" and then go on to belittle people who are "ignorant in science". What a clown. This thread just gets better and better. Gandalf61 (talk) 09:55, 12 February 2013 (UTC)[reply]
Is that the best you can do? No reasoned argument eh? You must be at last beginning to sense I must be right then. Wickwack 120.145.170.52 (talk) 11:01, 12 February 2013 (UTC)[reply]
Per mass-energy equivalence and conservation of energy mass is most certainly energy, no "conversion" between them is needed. For instance, one doesn't need to convert the energy of gluons to obtain the proton's rest mass or its total energy. The possibility of particle masses being wave energy was recognized by physicists such as Max Abraham before Einstein actually. I will add that in spite of the success of these principles, physicists are often either unclear or uncertain as to why the equivalence between these basic forms of energy exists, which is part of the reason why they continue to pour whopping amounts money into continued research. -Modocc (talk) 13:31, 12 February 2013 (UTC)[reply]
The first article you linked is about the E = mc2 equality. It's not about saying that mass is the same as energy, it about saying that if you convert a mass (the article gives nuclear fission as an example, but other methods of assembling and dissassembling particles are always conversions), then this is how much energy you can convert the mass into. It's about conversion. The second article you linked doesn't alter that, it just says the the convservation of energy principle had to be reviewed in light of the discovery that it can be converted to mass and back again. It's about conversion. However, for some theorists, a packet of energy (suitably sized and packaged up, not with brown paper but perhaps with some "string" http://en.wikipedia.org/wiki/String_theory) can manifest itself as a mass. Regardless, how does all this talk of mass-energy equivalence, gluons, and whatnot help the OP? How is it relavent to his questions about whether sound, light, and heat are forms of energy? How does it help him understand whether or not the energy in a radiation is related to frequency? Seems to me it has no relevance and may be for the OP a red herring. When one calculates how much energy is in a given sound wave, E = mc2 doesn't come into it. When you calculate the power of a light beam or radio wave, E= mc2 doesn't come into it. Wickwack 60.230.208.11 (talk) 14:51, 12 February 2013 (UTC)[reply]
From the first article I linked to: "According to the theory of relativity, mass and energy as commonly understood, are two names for the same thing, and neither one is changed nor transformed into the other." Apparently you are definitely not reading these articles carefully (Disclaimer: I didn't write these articles and its best to research the citations for a better understanding, and I've done my own reading of books and papers over the years so I do know what I'm talking about), and you could bring this up on the articles' talk pages and provide wp:RS for your view, which, in principle, dogmatically pigeon holes mass and energy into distinct categories. -Modocc (talk) 15:14, 12 February 2013 (UTC)[reply]
With the aid of Edit|find... sure enough, "According to the theory....: is right there under the heading Conservation of Mass and Energy, and is the "packet of" idea I mentioned (the article uses the phrase "a mobile form of mass) - I wasn't as dogmatic as you claimed. But you haven't answered how does this help the OP? How is it relevant to the OP's questions? Wickwack 60.230.208.11 (talk) 15:35, 12 February 2013 (UTC)[reply]
Of course masses are mobile and to be somewhat clearer, these photon packets are packets of mass-energy. As to why this helps the OP, it should be obvious, because the question's title is "Forms of energy", which could be rewritten without any loss of meaning as "Forms of mass-energy". -Modocc (talk) 15:50, 12 February 2013 (UTC)[reply]
Iranian and Iraqis anti-aircraft air defense
How does the present Iranian anti-aircraft air defense compares to the Iraqi anti-aircraft air defense at the beginning of the 2nd Gulf war? OsmanRF34 (talk) 14:14, 8 February 2013 (UTC)[reply]
I couldn't find any direct comparison, but this article is rather scathing of Iran's air-defence capabilities; "Iran has even more problems with its land-based surface-to-air missiles. Its only modern systems are short-range man-portable systems and some 30 short-range Russian TOR-Ms suitable only for point defense. Its other systems are 30 short-range Rapier fire units and 15 Tigercats of uncertain operational status. Its longer-range systems include roughly (154) U.S. IHawks, (45) Russian SA-2s, (10) SA-5s and a limited number of CSA-1 Chinese versions of the SA-2. All are obsolete."
Iraq in 1991 had spent a large amount of money on the best system that the Soviets (and the French to an extent) were willing to supply them with. By 2003, they only had whatever hadn't been destroyed in 1991. Although US Air Force general John W Rosa told journalists: "The Iraqi air defence system is one of the toughest, most complex systems that we see in the world... It's very capable. They're constantly working to improve it, and they have been.", Andrew Brookes, an air specialist at the International Institute for Strategic Studies in London said "It's rubbish".[9]Alansplodge (talk) 17:13, 8 February 2013 (UTC)[reply]
If Iran's anti-aircraft is as sophisticated as their new fighter jet,[10] then it may have come from the same oversized G.I. Joe kit. ←Baseball BugsWhat's up, Doc?carrots→ 07:30, 10 February 2013 (UTC)[reply]
Kite with extremely long string
If someone had a kite with a very, very long string attached, how high could the kite reach before some physical process prevented it getting any higher? --Dweller (talk) 14:41, 8 February 2013 (UTC)[reply]
The current record may be 13,600 feet [11] but I don't think any physical process limits it to that low. "Go fly a kite" and try to better the record yourself? In the U.S. you will need FAA clearance though. Rmhermen (talk) 15:25, 8 February 2013 (UTC)[reply]
At such altitudes, does it still work like a kite? bamse (talk) 16:58, 8 February 2013 (UTC)[reply]
Yes, but at some altitude there won't be enough air to create enough lift (airfoil effect) to overcome the weight of the string. 74.60.29.141 (talk) 17:05, 8 February 2013 (UTC)[reply]
The limit has to be the weight of the kite and the weight of the string versus the amount of lift it can generate. As the kite gets higher, the string gets longer - and therefore heavier. You might think that this means that you need a larger kite to lift the weight - but that increased area also increases the forces on the string because the drag force on the kite is proportional to the area. So a larger kite needs a stronger (and therefore heavier) string...and a bigger kite to lift that additional weight...and so forth.
The difficulty is that if you double the area of the kite, you double the amount of lift (and drag) that it has. You also double its weight and you have twice the tensile strength needed in the string - which demands a string with twice the cross-sectional area - which has twice the weight per unit length. Hence, making a kite larger doesn't increase the height it can fly at.
To solve that, you're going to need to consider varying the thickness of the string along its length. Near the kite, the string has to be strong enough to support it's entire weight - plus the drag forces on the kite. But close to the ground, it only has to be strong enough to counteract the drag force. So you could save weight by using a thicker cable up near the kite and less thick close to the ground...but there must be a limit to how much that helps.
Ultimately, you're going to start to find that the lessening of the density of the atmosphere would reduce the amount of lift that the kite can get...but it would also reduce the drag on it, allowing for a larger kite without needed a heavier string.
Arguably, the Space elevator idea is the ultimate kite-like thing - it would not be kept up there by the wind, but by the centrifugal force as the earth rotates. Issues of tether strength and the variation in thickness over altitude are a big question for that kind of structure.
[(edit conflict)] ~ However, there are aerodynamic effects on the string itself that can provide lift, as per ballooning spiders -which is still not well understood. I haven't checked this out yet, but you might want to read: Aerodynamics of Kites. 74.60.29.141 (talk) 17:17, 8 February 2013 (UTC)[reply]
In a place with air flowing quickly upward, like the a stationary warm-core storm, there should be enough lift on the string to overcome it's weight. StuRat (talk) 18:41, 8 February 2013 (UTC)[reply]
Gliders, hang gliders and paragliders are like kites without string. A very lucky paragilder reached 32,600 feet due to cloud suck and lived to tell the tale. World altitude record for hang gliding is 38,800 feet, although this was a balloon launch not an ascent from ground level. World altitude record for gliding is 50,699 feet. Gandalf61 (talk) 17:22, 8 February 2013 (UTC)[reply]
The problem with gliders and such is that once they are high enough to get above the beneficial "slope lift" effects - they are entirely reliant on upwelling air to get their altitude. Significantly, once they are high enough, the effect of the wind is zero - the speed of the glider relative to the air is all that matters.
A kite, on the other hand, is tethered to the ground - so it can use the speed of the wind to gain altitude. It's speed relative to the air is whatever the wind speed is. But it's ability to make use of upwelling air is limited for the exact same reason...if it doesn't happen to be in an upward current, there is nothing that can easily be done to fix that. So, in principle, a kite can do better than a glider...especially if there is abundant wind and an updraft that happens to be where the kite is situated...but eventually, the weight of the string outweighs that advantage. SteveBaker (talk) 20:03, 8 February 2013 (UTC)[reply]
So, any idea of a reasonable maximum altitude? --Dweller (talk) 20:06, 9 February 2013 (UTC)[reply]
That depends on the string's strength-to-weight ratio more than anything else. 24.23.196.85 (talk) 20:14, 9 February 2013 (UTC)[reply]
For two kites then, one with typical commercial strength-to-weight ratio, and one superduper one designed by NASA engineers in their spare time. --Dweller (talk) 20:17, 9 February 2013 (UTC)[reply]
What material should I use for the string in each of the two cases?24.23.196.85 (talk) 20:19, 9 February 2013 (UTC)[reply]
Just ran some calcs for your second case (the "superduper one designed by NASA engineers in their spare time"): with a string made of Kevlar (assuming that the string has a constant diameter, and that I didn't screw up anywhere in my calcs, which I might well have), the ballpark figure for the theoretical "highest height" would be over 80,000 feet! Yes, you read it right -- assuming my calcs are correct, you can literally"send it soaring up through the stratosphere, up where the air is clear"! Of course, you'd first have to get permission from the FAA, because this would be a major aviation hazard... 24.23.196.85 (talk) 21:04, 9 February 2013 (UTC)[reply]
The FAA doesn't have much authority over Dunstable Downs (see parag 4), but I'm sure that someone would be annoyed if I deployed an 80,000 ft vertical string without notifying them. Not least the airfield that lies immediately below the popular kite-flying venue! --Dweller (talk) 08:48, 10 February 2013 (UTC)[reply]
So, 80,000 feet seems to be the max, determined by the weight of the string and the amount of lift still in the atmosphere? That's a pretty long piece of string. Thanks guys. --Dweller (talk) 08:48, 10 February 2013 (UTC)[reply]
Yes, and it would cost you a lot more money than a "twopence for paper and string" (not to mention that the kite, too, would have to be made of synthetic fabric in order to survive the strong winds and turbulence in the upper atmosphere); also, even with a Kevlar string that's only 1 sq. mm. in cross-section, 80,000 feet of it would weigh over 75 pounds. (Not to mention all the other logistical challenges that such a record attempt would entail.) 24.23.196.85 (talk) 01:26, 11 February 2013 (UTC)[reply]
It's a lot worse than that. Unless the string is dead vertical and there's no wind loading (and if there's no wind you can't fly the kite), you get a catenary loading effect in the string. Visualise an inelastic string that is slightly curved due to gravity, wind loading, or both. If the centre of the string length is displaced x metres, the ends are drawn towards each other in order to keep the same string length. The distance each end is drawn in is very much smaller than x and there is in consequence a considerable force mulitiplication as with any lever where the load moves a larger distance than the lever end. So, while the string may weigh only 75 pounds, the pull on the kite will be much much larger. In practice this will force you to use a much longer string so it can curve considerably - this will directly increase the weight of the string. Wickwack 121.221.38.48 (talk) 11:18, 12 February 2013 (UTC)[reply]
Darn, I knew I missed something... 24.23.196.85 (talk) 03:15, 13 February 2013 (UTC)[reply]
If there were a number of kites along the string as in the picture on the right then one needn't worry about the weight, and the air going in different directions at different heights might actually help by reducing the horizontal force as well. Don't know if anyone has tried that though. Dmcq (talk) 14:00, 12 February 2013 (UTC)[reply]
Symbol of mass number
The symbol of atomic number is Z and the reason for this is beautifully mentioned in the article. The symbol of mass number is A. What is the reason behind using A as the symbol for mass number ? Show your knowledge (talk) 18:08, 8 February 2013 (UTC)[reply]
Electrons are negatively charged and nucleus is positively charged, we also know, unlike charges attract each other. Why don't electrons come and stick to the nucleus of the atom ? Want to be Einstein (talk) 18:17, 8 February 2013 (UTC)[reply]
They are in orbit. Just as the Earth would fall into the Sun, due to gravity, but doesn't, because of it's momentum, the same is true of electrons. There is nothing to slow them down, so they just keep on going. When a stray electron "gets lucky" and slams into a nucleus, bad things happen, like it hitting a proton and fusing into a neutron. However, the vast amount of empty space between the electrons and nucleus, compared with the size of electrons, makes such collisions extremely rare, a bit less so where the electrons are moving faster, like inside giant stars. StuRat (talk) 18:55, 8 February 2013 (UTC)[reply]
I remember two things that I have read somewhere on Wikipedia but I am not able to find them: First, electrons don't orbit the nucleus unlike planets around sun. Electrons have random motion instead of well-defined circular motion. Second, the reason StuRat mentioned is not the correct answer to my question. Want to be Einstein (talk) 19:13, 8 February 2013 (UTC)[reply]
Here you get into the classical model versus quantum mechanics. I gave you the classical model. Look at Jayron's answer below for more of the quantum mechanics model. StuRat (talk) 19:24, 8 February 2013 (UTC)[reply]
Actually, your classic model doesn't work for the following reason: An electron is charged whereas the Earth is essentially neutral. Electrical charges emit radiation when the accelerate (under classical physics). See Larmor formula. An orbit is an acceleration (all turns are accelerations); so an electrically charged particle should be radiating energy constantly as it turns its orbit around the nucleus, this loss of energy should cause the electron to spiral into the nucleus. The reason the earth doesn't do this around the sun is that the earth is electrically neutral, and so does not radiate when it is accelerated, and so can maintain a steady-state orbit. The fact that an electron has an electrical charge means that if it really was a particle in an orbit, it would either need a constant resupply of energy or it would spiral into the nucleus. The fact that this doesn't happen may be one of the key impetuses for developing quantum mechanics to explain how the atom was able to maintain a steady state despite this. --Jayron32 19:34, 8 February 2013 (UTC)[reply]
Does a single electron radiate energy when it turns, as when deflected in a CRT screen ? Also, the quantum mechanics explanation seems to come down to "we don't why, just accept it", as you put it below, so that's not really any better, just more complicated. StuRat (talk) 19:38, 8 February 2013 (UTC)[reply]
Yes, electrons radiate energy when they are accelerated; when the electron fired in a cathode ray tube hits the phosphor, it will do so with slightly less energy than it would be predicted if the electron weren't charged. That's a principle which has been known since almost before people even knew that electrons existed. The Larmor principle and the first description of the electron date to the same year (1897). However, the better way to understand the principle is radio. Radio waves (photons, or light energy) is generated fluctuating electrical current. The source of the radio waves is the energy given off by variations in electrical current; you speed up and slow down electrons, and they shed radiation as a result. By controlling how you vary the current, you control how the radio waves are varied, and you can transmit information that way. That works exactly because accelerating electrons give off radiation. If the electron in an atom were in orbit, it would be under a constant acceleration, and would be similarly radiating. That it isn't is why we can safely say that the electron isn't actually orbiting. --Jayron32 19:48, 8 February 2013 (UTC)[reply]
Oh, and the "just accept it" thing isn't an admonishment to claim that QM can't be understood; it's that the implications of QM cannot be visualized. The problem is that people want to have a picture, some analogue they could create with shapes and objects they are familiar working with. You simply cannot visualize an atom like this; the problems with accepting QM principles is that you have to abandon the need to have a visual model which can represent them. So, I'm not saying "just accept it" to mean "humans cannot understand this", I'm saying "just accept it" to say "there's nothing in the way that classical objects work that can be a good analogue of this". Since, as humans, our entire sensory experience is with classical objects, there is literally no convenient way to describe a picture which correctly displays QM principles. So you have to accept what the calculations and laws tell you without having a picture to go along with them. --Jayron32 19:54, 8 February 2013 (UTC)[reply]
So how does QM explain the underlying reason why a cloud with no moving parts has momentum ? Does it come down to anything more than "that's what we observe" ? StuRat (talk) 20:02, 8 February 2013 (UTC)[reply]
That's the point. Concepts like "clouds", "moving", and "parts" are defined classically. When you say it's better to picture an electron as a "cloud", that's still depending on drawing an analogue to a classical object, and atomic level physics simply doesn't have those analogues. Electrons aren't balls, and they aren't clouds. The cloud visualization is better (because it captures the idea of nonlocalizability) but it has its own major fault in that such a steady state cloud cannot also have momentum, at least if your depending on what your experience tells you a cloud is. Quantum scale physics has all of these problems in meshing with human experience. The entire concept of wave-particle duality is a completely different sort of problem, but it's of the same ilk. A wave is a type of movement. A particle is a thing. So how can a thing be a type of movement also? One particularly bad explanation of wave-particle duality is that light, for example, chooses which set of properties it has based on the application. That's wrong headed. Light behaves the exact same way all the time in all situations. We can construct situations where modeling light as a particle works better, and we can also construct situations where modeling light as a wave works better; but neither model really captures what light is. There is nothing your senses experience that allow you to be able to visualize what light is. The best we can do is fudge together some classical concepts like "waves" and "particles": it's a shortcoming of human perception that's the problem. What we observe is a set of properties: the ways in which atoms interact with each other, the ways in which molecules form and behave, the ways in which electrons around a nucleus absorb and emit energy, etc. We run experiments and get data from those experiments. If we try to match that data with the predictions made by the equations of classical physics, it just doesn't work. That's what quantum physics does for us: it gives us a new set of equations that correctly match the experimental data of how electrons work and also correctly match the experimental data for everything else as well. As I mention below, the reason we keep the old equations around for the non-atomic stuff is that they're more convenient to work with, not because the quantum equations don't also work. --Jayron32 20:18, 8 February 2013 (UTC)[reply]
But don't you see a problem with just saying "that's the way it behave because that's what the equations say" ? This comes up in string theory, where we can come up with any of several set of equations, all of which match observations to the same degree. So, which is right ? Are we just guessing here ? StuRat (talk) 21:17, 8 February 2013 (UTC)[reply]
You're starting to edge up against a complete misunderstanding of what science is, Stu. Science doesn't tell us what is right, science tells us what works. The fact that there are multiple solutions that produce the same result is why things like string theory are open areas of research and exploration in science. We don't know the answers to the unanswered questions in physics, and a big part of that comes down to working out which of any of a number of competing theories (if any of them so far proposed) is more useful in explaining the current inadequacies in our existing models. "All models are wrong, some are useful" is an important aphorism here in understanding this. We're not trying to decide absolute rightness, we're trying to come up with more and more accurate models; but no model will ever be complete. The ones we have now are fantastically useful and accurate, much more so than Newtonian physics we used to work with (and still do work with where Newtonian physics agrees with the better, but more complex and esoteric, models). As simply as I can say it is this: the more modern theories (like QM and GR) which have gained broad acceptance are better than classical physics because the explain more phenomena in better detail, and match observed data more accurately than classical physics does. There are areas of open exploration in science (like the String Theory you keep bringing up) which are attempt to add another layer of accuracy and precision to our understanding, but which are not yet fully fleshed out nor fully accepted, because of the problems of the "several sets of equations which all match observations to the same degree". It is not a simple issue, there's thousands of physicists right now that are exploring those avenues and trying to resolve those problems you note. Most of them will go down dead ends, but perhaps someone working now will be able to add some positive confirmation to something like string theory, or another unifying theory, or maybe come up with something else. Unlike things like quantum mechanics and general relativity, these other ideas are just too new and too untested to produce anything like universal acceptance. One thing that is for certain, however, is that the new models will still need to agree with observable data at all scales and for all phenomena; if they don't they're not very useful. If they do, but don't add to the corpus of explained phenomena, then they're also not very useful. Useful new models are only those which match everything we already know AND which help to explain something we can't already explain. --Jayron32 22:35, 8 February 2013 (UTC)[reply]
When QM was new, it didn't predict everything, like the behavior of gravity, which was better explained by the older GR. Yet QRQM was generally accepted, because it explained other things better. StuRat (talk) 22:58, 8 February 2013 (UTC)[reply]
Assuming you mean GR in your last sentence, YES. --Jayron32 23:26, 8 February 2013 (UTC)[reply]
Nope, I meant QM. StuRat (talk) 00:07, 9 February 2013 (UTC)[reply]
Then the answer is still YES. --Jayron32 00:17, 9 February 2013 (UTC)[reply]
Agreed, thus refuting your statement that "Useful new models are only those which match everything we already know AND which help to explain something we can't already explain". StuRat (talk) 00:21, 9 February 2013 (UTC)[reply]
No, both Quantum Mechanics and General Relativity match everything we already know from classical physics (that is, neither of them contradicts the experimental results that classical mechanics also matches), but both theories also correctly predict things that classical physics gets wrong. That's why they are useful: They expand upon and replace simpler theories, and get more correct than the simpler theories do. If, for example, Quantum mechanics only gave us the exact same results as classical physics, and didn't give us better explanations for the things classical mechanics got wrong, it wouldn't be useful. If quantum mechanics could better explain some things that classical mechanics got wrong, but it was also wrong where classical mechanics was correct, it ALSO wouldn't be useful. That they agree in areas where observation confirms that classical mechanics was correct, and that quantum mechanics ALSO adds entire new explanations for physical phenomena that classical mechanics gets wrong (like the aformentioned problem of the electron that started this thread) is why QM is a generally accepted theory. It agrees with all of the observations classical mechanics agreed with, and adds a whole new set of observations that classical mechanics couldn't correctly predict. You can replace the words "quantum mechanics" with "general relativity" and the same statements hold true. --Jayron32 00:31, 9 February 2013 (UTC)[reply]
I wasn't talking about classical mechanics here. I was refuting your statement that "Useful new models are only those which match everything we already know AND which help to explain something we can't already explain" by giving the example where QM was a useful new model, despite not matching everything we already knew (from GR). Also, a model which explains nothing new, but is simpler, may also be useful. This comes up in reverse with string theories which want to add more and more dimensions, with little benefit. StuRat (talk) 00:21, 9 February 2013 (UTC)[reply]
The fact that GR and QM don't play well together is already well established and not in dispute. I already cited it as one of the great unanswered questions in physics. So, you're not adding anything surprising to the physics canon by noting that. --Jayron32 00:49, 9 February 2013 (UTC)[reply]
And I'm not trying to do that, am I ? I just used it to refute your statement, as I've said twice now. StuRat (talk) 06:43, 9 February 2013 (UTC)[reply]
The answer you gave is like explaining that the Earth is flat and stars are holes in a black cloth - that would be the "classical model". Your answer has been known to be incorrect for almost a century. 88.112.41.6 (talk) 19:39, 8 February 2013 (UTC)[reply]
It's not that simple. Quantum mechanics seems to work better on a small scale, and the classical mechanics approximation of relativity or relativity itself are better on a large scale. How to mesh them together is a problem that still confounds us. Things like electrons are right at the cusp of the two models, sometimes behaving like a probability cloud, and sometimes like a particle. So, relativity (and the classical mechanics approximation of relativity) isn't strictly right or wrong, and the same is true of quantum mechanics. StuRat (talk) 19:44, 8 February 2013 (UTC)[reply]
Your explanation is known to be incorrect. It violates basic laws of physics (conservation of energy and momentum). You may want to consider if your frequent contributions to the reference desk would be more valuable if you could recognize and admit when you give a wrong answer. This is not the first time I see you doing this. 88.112.41.6 (talk) 19:50, 8 February 2013 (UTC)[reply]
Actually, Stu, we do understand perfectly well how QM and classical mechanics mesh. QM laws reduce just fine to classical laws in the limits of measurements; they scale perfectly. There is no magical moment when an object stops obeying quantum rules and starts obeying classical rules; there's a continuum where the difference between the predictions made by classical physics and quantum physics decreases as the scales increase. That is, as size scales become larger, the quantum laws predict the exact same properties as the classical laws do. The entire universe can be accurately modeled entirely using quantum mechanics. We keep classical physics around because the math of quantum systems is a bitch, and if we can get the same results with the easier math, then why not? But it simply isn't true at all that quantum mechanics doesn't work on large scales. Only the converse is true: Classical physics doesn't work on small scales, but quantum physics works on all scales. That's why it's a better overall theory. --Jayron32 20:00, 8 February 2013 (UTC)[reply]
I don't agree that QM works on all scales. Here's a source which backs me up: [12]. And the opening sentence in our article says "Quantum mechanics (QM – also known as quantum physics, or quantum theory) is a branch of physics dealing with physical phenomena at microscopic scales, where the action is on the order of the Planck constant". Now, string theory attempts to unify both classical mechanics and QM, but it has it's own problems, such as having many variants, all of which are untestable. StuRat (talk) 20:06, 8 February 2013 (UTC)[reply]
And now, rather than admitting to being wrong, you go into obfuscation mode. The original question was why don't electrons go into the nucleus. Nothing to do with "scales". You gave the exact anti-answer - a known incorrect answer that was thrown out the window exactly because it was discovered that it leads to electrons going into the nucleus, making all atoms in the universe disappear in a fraction of a second. Your habit of doing this is unfortunate given the volume of text you write on the reference desk. 88.112.41.6 (talk) 20:24, 8 February 2013 (UTC)[reply]
Put up or shut up. I've provided links supporting my point. If you disagree, let's see your proof, not personal attacks. StuRat (talk) 20:29, 8 February 2013 (UTC)[reply]
Holy shit, that's your source? Don't mind if I don't believe a word of what's written on some hand-made website. I've not applied the Crackpot index to it, but my sense tells me it'd be ridiculous. Seriously Stu, if you're going to make a claim like that, you're going to need a better source. The same Wikipedia article on quantum mechanics you cite also states "Predictions of quantum mechanics have been verified experimentally to an extremely high degree of accuracy. According to the correspondence principle between classical and quantum mechanics, all objects obey the laws of quantum mechanics, and classical mechanics is just an approximation for large systems of objects (or a statistical quantum mechanics of a large collection of particles). The laws of classical mechanics thus follow from the laws of quantum mechanics as a statistical average at the limit of large systems or large quantum numbers.[35] However, chaotic systems do not have good quantum numbers, and quantum chaos studies the relationship between classical and quantum descriptions in these systems." (bold mine) I mean really Stu, if you're going to make silly claims like this, don't quote a source (the Wikipedia article) that directly contradicts you. Seriously, read the entire article next time. You might actually learn something new. --Jayron32 20:27, 8 February 2013 (UTC)[reply]
What's that opening line saying about Plank distances, then ? Also, out Theory of everything article says it's "... a theory that would unify or explain through a single model the theories of all fundamental interactions and of all particles of nature: general relativity for gravitation, and the standard model of elementary particle physics — which includes quantum mechanics — for electromagnetism, the two nuclear interactions, and the known elementary particles". So, if you're claiming that QM already does that, then why do we need a ToE ? StuRat (talk) 20:32, 8 February 2013 (UTC)[reply]
Planck distances are the distances when the classical model stops working, which is why we need the quantum model to work there. It has nothing to do with the upper limit for quantum models. The quantum models work on all scales. Also, QM doesn't do everything. QM doesn't have a good explanation for gravity, which is why we need a theory of everything. Gravity is very well modeled by general relativity in the sense that general relativity makes mathematical predictions about gravity that are borne out by experimental data, and which classical explanations of gravity do not. The other fundemental forces are well modeled by quantum mechanics: that is QM makes predictions about how forces like the nuclear forces and electromagnetic forces behave, and those predictions are borne out in the experimental data. The problem is that we have two post-classical theories, each of which very accurately matches experimental data in their own domains (and each does a better job in those domains than classical physics does) which appear to be incompatible with each other. --Jayron32 20:47, 8 February 2013 (UTC)[reply]
It sounds like the lede to our QM article needs changing, then, since it really doesn't say that now. Also, since gravity is critical to understanding macroscopic behavior, if QM doesn't explain it, then QM does have an upper size limit, be that the Plank length or something else. StuRat (talk) 21:22, 8 February 2013 (UTC)[reply]
Furthermore, Sting theory isn't trying to marry classical mechanics and quantum mechanics. It's trying to marry general relativity and quantum mechanics. GR and QM are two different non-classical theories that operate in different domains (GR models gravity, while QM models other forces) --Jayron32 20:31, 8 February 2013 (UTC)[reply]
OK, I was a bit sloppy in equating GR with classical mechanics, but they do rather go together, as GR does work with large objects, like stars and planets, and not-so-much with wave probability functions from QM. I like to think of GR as tweaking classical mechanics, while QM tosses it out the window entirely. StuRat (talk) 20:39, 8 February 2013 (UTC)[reply]
It's a far cry from a mere "tweak" of classical gravity; it also throws classical gravity out the window. Classical gravity has two problems 1) it propagates instantaneously instead of at the speed of light and 2) classical gravity is a real force, whereas in GR gravity is a pseudoforce akin to centrifugal force. That is, General Relativity says that gravity isn't a force at all; it's a geometric warping of space time that gives the observer the 'illusion' of a force. This idea that spacetime can be warped isn't a mere tweak, its the central aspect of GR and is completely and totally out of the realm of classic physics. --Jayron32 20:53, 8 February 2013 (UTC)[reply]
I call it a tweak since it seems mostly to just be another way to visualize gravity. The results are similar, whether you think of space-time as warped or explain orbits with classical mechanics. The speed of light considerations are where more of a difference is observed between the two. StuRat (talk) 21:10, 8 February 2013 (UTC)[reply]
If by "tweak" and "just another way to visualize" you mean "a complete and total rewrite of every aspect of our understanding of how the universe works on the most fundamental levels" then you may be closer to the truth. Seriously Stu, I say this out of love, because I don't want to see you embarrass yourself further along this line of thinking, but you're like a walking, talking personification of the Dunning–Kruger effect. You started with a completely incorrect premise, and rather than concede that you continue to make statements which, bafflingly, are progressively wronger and wronger. GR is not another way to visualize gravity. GR is a fundamentally different way to understand the entirety of physics and motion and kinetics and dynamics and velocity and acceleration and time and space. It has it's own mathematics, it's own geometry, there's absolutely nothing that Newton would look at and say "Yup, that's basically what I said". The results as far as gravity is concerned are not "similar", excepting in the classical limit. At an approximation, the kinds of behaviors that GR predicts about how, say, an apple will move as it descends towards the ground are identical to what classical physics says; this is analogous to the way in which quantum mechanics and classical mechanics converge to the same mathematical results when you get to large objects. However, that doesn't mean that the kinds of differences between the predictions of GR and classical mechanics makes are just minor tweaks. Classical gravity is way off and doesn't correctly model lots of behaviors correctly at all; things like black holes and gravitational time dilation and Frame-dragging and the Geodetic effect which classical gravity gets wrong in the exact same way that classical mechanics gets the atom wrong. There's just no intellectually honest way to say that General Relativity was merely a change of perspective or a tweak to classical gravity: it's a fundamentally different way to look at the universe, and it makes different predictions that match experimental results in ways that classical gravity just does not. --Jayron32 22:16, 8 February 2013 (UTC)[reply]
Look at my example, of planetary orbits. How does GR predict different orbits from classical mechanics ? Newton would definitely say "that's just another way to look at it", not, "I was wrong". It's only when you get far from our everyday experiences that GR predictions are far removed from those of Newton. StuRat (talk) 22:52, 8 February 2013 (UTC)[reply]
Well yes, in general, but even with this GR predicts the correct orbit of Mercury but Newtonian physics does not.--GilderienChat|List of good deeds 23:23, 8 February 2013 (UTC)[reply]
Newton gets the orbits of planets wrong by incorrectly predicting the way that they precess. The observed apsidal precession of the planets does not in accurately match the behavior predicted by simple classical mechanics (Newton and Keppler and all that jazz), you need general relativity to predict it correctly. It's wrong for all planets, but noticably so for Mercury; the incorrect predictions about Mercury's orbit are large enough that they were noted before Einstein was even born. No one could explain it until General Relativity created a model that fit perfectly. If you used a GPS in the past week, if it were not for General Relativity, said GPS would have been off by miles in telling you what your position is: based on classical understanding of gravity, your position cannot be accurately calculated using GPS, because classical gravity has no means to calculate Gravitational time dilation which must be taken into account given the distance the GPS sattelite is from the surface of the earth. Newtonian gravity can't explain Gravitational lensing, something astronomers work with every day. Also, no one said Newton was wrong. As I said above, science isn't looking for 'right'. Science is looking for 'useful'. Newton's theories are still, to this day, very useful: most of the stuff you do on a daily basis that may require basic physics calculations, such as the speed your car moves, or the trajectory of a thrown object, or any number of other common calculations, are perfectly well predicted by Newton's theories. Where they fall short, the more complete General Relativity steps in to help expand the corpus of applications, for example the two shortcomings I noted above. Now, General Relativity also predicts the same things about the speed of your car and the flight of a thrown object that classical physics does. It's just silly to do all that math to get the same answer in those sorts of applications. --Jayron32 23:25, 8 February 2013 (UTC)[reply]
I'd still call all that "tweaking". For an example of "a complete and total rewrite of every aspect of our understanding of how the universe works on the most fundamental levels", I'd go with the ancient Greek model of the "elements" being earth, air, fire, water, and ether, each corresponding with a perfect solid, versus our current understanding of the elements. StuRat (talk) 00:13, 9 February 2013 (UTC)[reply]
Apples and oranges. There's no fundamental way you can call the Platonic elements "science". However, the differences that General Relativity wrought on our total understanding of physics is akin to the difference between Dalton's atomic theory and the modern Atomic orbital model of the atom. That is, what Dalton said the atom looked like and what modern Quantum mechanics thinks the atom looks like is roughly akin to the difference between what Newton said the physical universe worked like and what GR says of the same. --Jayron32 00:23, 9 February 2013 (UTC)[reply]
(edit conflict) Damn good question. Since the structure of the atom was first elucidated by Rutherford in his gold foil experiment, that was a major, central controversy with his nuclear model of the atom. They should crash into the nucleus under the principles of classical physics, and there's nothing in classical physics which adequately explains why they don't. Indeed, that problem is one of the keystones which brought classical physics (as a means to describe atomic-level phenomena) down and led to the development of more advanced, modern understandings of the physics. Now, the first thing you need to do is to take the image of the little electron orbiting the nucleus like the earth does the Sun and put it in the same part of your mind that you keep Santa Clause and the Easter Bunny: it's a nice little story we tell kids, but its a total fiction. An electron is not a little ball. Trying to describe an atom as you would describe any other object is always problematic, but if you must create a picture in your mind, it is better to think of an electron as a cloud instead of a point. The cloud exists over a volume of space described by a wave function which describes both the shape and density of that cloud over space. Now, that cloud is imbued with certain properties that must be conserved; that is no change to that cloud can eliminate said properties. The electron cloud has, for example, momentum. Now, this momentum exists spread out over the whole cloud, and cannot be localized to any one point in the cloud (I know, this makes no sense when you try to picture it, but it doesn't need to make sense to be true. This is one of those things you have to be able to accept without a visual representation). Any changes to the cloud, such as its size, must conserve this momentum. If the cloud were entirely compressed into the nucleus, then this momentum would be localized into a single point. The fact that this kind of localization of momentum is impossible is enshrined in the uncertainty principle, which is a cornerstone theorem of modern physics. So the reason that the electron never crashes into the nucleus is that, if it did that, its momentum would be localized in one location, and the uncertainty principle says that you can't localize momentum in that way. The uncertainty principle says any particle cannot simultaneously have a precisely defined location and momentum; since all electrons have a precisely defined momentum, they can NEVER have a precisely defined location. So electrons cannot "crash" into the nucleus.
Another way to think of it is this: there is a force holding the electron to the nucleus in the same way that there is a force holding a magnet to, say, your refrigerator. Now, there is not any loss of energy as the magnet holds fast to the refrigerator. The magnet has potential energy, but so long as the magnet does no work, then it will remain stuck to the refrigerator forever without losing any of that potential energy. Work only happens when a force moves something; forces that don't move objects don't expend any work, and no energy is "used up". In almost the exact same way, the electron is "held fast" to the nucleus by the electrostatic force, but no energy is lost because the electron does no work in remaining there. Here's where the classical explanation and modern explanation diverge: if you're picturing the electron as a little ball, there's no way for it to remain in motion under such a force and do no work at all. Such a ball should spiral into the nucleus, shedding potential energy the whole way in. If, however, you think of the electron as a steady-state cloud it just remains in the same state forever, and does no work. Now, the mindfuck here is that you need to accept that this steady state cloud still has momentum without actually having any localized parts which are in motion. There's no convenient way to make that work except to say "just accept it". --Jayron32 19:00, 8 February 2013 (UTC)[reply]
Well technically in hydrogen it sort of does - the 1s electron is in an orbital which is a sphere shape and superimposed over the proton, which is also technically a sphere of probability. So under some observations, it will appear to be "inside" the nucleus.--GilderienChat|List of good deeds 20:20, 8 February 2013 (UTC)[reply]
And keep in mind that the electron has half a spin all by itself, in addition to the full spin of orbiting around the nucleus. So the final result is either a spin and a half if aligned, or half a spin if opposed and no other result can ever be measured. Hcobb (talk) 20:23, 8 February 2013 (UTC)[reply]
As is multiply explained above and in the linked articles, "spin of orbiting around the nucleus" is a load of completely disproven nonsense, no? DMacks (talk) 22:06, 8 February 2013 (UTC)[reply]
Well as I read it, the "cloud" does have momentum and this would thus be either in one direction or another (as a non-zero vector quantity must be) and so these effects will be observed.--GilderienChat|List of good deeds 23:35, 8 February 2013 (UTC)[reply]
self explanatory... μηδείς (talk) 20:43, 8 February 2013 (UTC)[reply]
Well, two things 1) It isn't (the search you gave gives LOTS of threads that mention the words "electron" and "nucleus" without addressing this specific problem) and 2) If it is a commonly asked question, that's because of what I explained above: things like atoms, electrons, light, nuclei, etc. etc. don't have analogues to what you can sense and experience in the world. There is literally no shape, object, behavior, or concept about how you experience the world with your five senses which correctly and accurately models quantum behavior. That is, the predictions you would make about an electron if you say "an electron is like FOO", where FOO is literally anything you have ever experienced, always breaks down. Some representations work in some situations, which is why we use them in some explanations, but no such representation works all that well. So, you need to rely solely on two things A) the data from experiments and measurements and B) the predictions of the equations of quantum mechanics. As long as A = B, it's a highly useful theory. The fact that it's also not a theory that lends itself to easy visualization doesn't invalidate it; but it does make it very hard for people to wrap their heads around, since we all learn by analogy; we try to connect something we're learning to something we already know. QM resists that sort of analogy, which is why it is so confusing, and why people have a hard time wrapping their heads around it. --Jayron32 21:11, 8 February 2013 (UTC)[reply]
I wasn't challenging any prior answers, just pointing out the same question was asked about a month ago and several other times I remember. A search on "fall into the nucleus" gets you lots of prior answers, only a few of which are links to Dr Who episodes. μηδείς (talk) 02:53, 9 February 2013 (UTC)[reply]
I wanted to comment on a few things, but didn't want to insert my comments into the above mess. So I'm putting all of them here.
First, why the heck are we debating whether quantum mechanics works for classical systems? It does, period, end of story. In fact, we have an article about this: correspondence principle, which gives plenty of examples. In any undergrad QM class, you get to prove ad nauseum how your quantum result for various problems reduces to the classical result in the limit of high energies and large sizes. If it doesn't, then you did something wrong, and have to redo the problem. QM is incompatible with general relativity because the former cannot describe strong gravitational fields, not because it can't describe large sizes. It's quite easy to describe weak gravitational fields in QM: just introduce a classical 1/r gravitational potential. Strong gravitational fields usually arise at very small scales, like the singularity of a black hole or the very early universe, not at large scales like that of the solar system.
Second, in the classical model where the electron orbits a positive nucleus, the electron would radiate energy and crash into the nucleus. As Jayron noted, this fact is expressed in the Larmor formula. This, again, is indisputable. Bohr model#Origins describes the historical significance of this (although strangely, Rutherford model doesn't):
"Rutherford naturally considered a planetary-model atom, the Rutherford model of 1911 – electrons orbiting a solar nucleus – however, said planetary-model atom has a technical difficulty. The laws of classical mechanics (i.e. the Larmor formula), predict that the electron will release electromagnetic radiation while orbiting a nucleus. Because the electron would lose energy, it would gradually spiral inwards, collapsing into the nucleus. This atom model is disastrous, because it predicts that all atoms are unstable."
Third, here's an alternative answer to the OP's question, which is equivalent to Jayron's in the sense that both use the same underlying theory. In quantum mechanics, the electron is represented as a wavefunction. The physical significance of the wavefunction is that if you try to measure the electron's position, the square of the wavefunction at any point is equal to the probability of finding the electron at that point. Schrodinger's equation tells you what the wavefunction can possibly look like. So, why is it not possible for the wavefunction to be extremely densely concentrated around the nucleus? Because such a wavefunction doesn't satisfy Schrodinger's equation. You can see every function that does satisfy the equation; they're called atomic orbitals. Note, however, that the squared wavefunction is only a probability distribution, and even very close to the nucleus, it doesn't fall to 0. You might ask, if this is the case, why electrons don't sometimes fall into the nucleus. The answer is that they do, and this process is called electron capture.
Finally, contrary to what Jayron said, electron clouds have no momentum. This is because atomic orbitals are stationary states, and the expected value of the momentum operator is always 0 in any stationary state. The expected value of the momentum squared, however, is non-zero, so electrons have non-zero kinetic energy. --140.180.247.198 (talk) 00:35, 9 February 2013 (UTC)[reply]
Thanks for correcting me on the momentum/kinetic energy thing. I confused the terms above in my initial explanation. However I am still pretty confident in the uncertainty principle implications of having an electron located in the nucleus. I did some more digging just now to check up on the other aspects of my explanation, and found this explanation from the University of Illinois physics department, which may be helpful towards answering the initial question. --Jayron32 00:47, 9 February 2013 (UTC)[reply]
Well, I am not a expert of Quantum Mechanics and the above discussion is completely based on Quantum Mechanics. I have thought a answer to my question but I think that is wrong. "Suppose, an electron gets attracted to the nucleus, it falls towards the nucleus, coming to a lower orbit and then to other lower orbits between the electrons's own orbit and the nucleus. We know, for an electron to come to a lower orbit, it would have to emit photon. The attraction of nucleus for an electron is not so strong that it will make an electron to emit photon. Therefore, the electron doesn't come to a lower orbit. This is why electrons don't fall in the nucleus". Am I right or wrong? Want to be Einstein (talk) 03:46, 9 February 2013 (UTC)[reply]
See my answer in this old thread. GilderienJayron32 made some similar points earlier in this thread. -- BenRG (talk) 06:02, 9 February 2013 (UTC)[reply]
Watch this video I know there is a way to freeze water in 5 seconds. But I do not know how to do it. After watching the linked video please tell me a way by which I can also make water freeze very quickly. Polar Bear25 (talk) 19:57, 8 February 2013 (UTC)[reply]
Well, there are four factors on how to freeze water quickly, in general (although this seems to be a case of supercooling, see answer below):
1) Have it as close to freezing temperature as possible, before you start.
2) Expose it to extreme cold.
3) Make the water particles as small as possible.
4) Lower the pressure.
So, if you spray a fine mist of high-pressure, near-freezing water through liquid nitrogen, while lowering the temperature, you should be able to freeze it considerably quicker than 5 seconds. StuRat (talk) 20:09, 8 February 2013 (UTC)[reply]
(ec) It's called supercooling. It's a trick in which purified water is cooled down very carefully below freezing point so that it gets no 'opportunity' to crystallize. If the water is disturbed, it then suddenly starts to form ice crystals. When you see the water freezing, its temperature does not actually decrease, but it increases because energy is released by the formation of strong bonds in the crystal. - Lindert (talk) 20:14, 8 February 2013 (UTC)[reply]
How can I do this as the person in the video did ? Polar Bear25 (talk) 03:57, 9 February 2013 (UTC)[reply]
There are plenty of videos on the web, e.g. this one. For best results use distilled or spring water, and leave in freezer for about three hours.--Shantavira|feed me 10:18, 9 February 2013 (UTC)[reply]
You are on the right path. Placing bottle in freezer water, but how will I freeze it. Polar Bear25 (talk) 11:09, 9 February 2013 (UTC)[reply]
Wash and rinse the bottles carefully, rinsing the last couple times with distilled water. You need to have the timing right, so they are below freezing, but not too much. Or the freezer temperature needs to be set just below freezing temp. Then, just shaking one should make it freeze. An alternative is to remove the lid. The lowered pressure may also trigger freezing, so it could freeze before you get the lid off.
However, note that there's considerable luck involved. Some stray particle which acts as a nucleation site for ice often gets into the water. So, many bottles of water would give you a better chance that some become supercooled, and, when some start to form ice, that's a good temperature indication, so, if you see others interspersed with the frozen ones which aren't frozen, then you could shake them to try to get a quick freeze out of them
One word of caution: Be sure to fill the bottles only maybe 2/3 full. The air gap is needed since water expands when it freezes, and needs someplace to expand into, or a glass bottle might shatter. StuRat (talk) 16:58, 11 February 2013 (UTC)[reply]
Space between atoms
In the article atomic spacing it says that the distance between atoms varies from a few angstroms in solids to "as large as a meter" in "outer space". Yet I asked a similar question before here and was told that the distance between the inner electron and the nucleus could technically be infinite. The wording in the article also implies that if the distance between 2 or more atoms can be measured then the distance between the electron cloud and the nucleus can also be measured. Is there a rule for determining the distance between the inner electron cloud and the nucleus of an atom?165.212.189.187 (talk) 20:38, 8 February 2013 (UTC)[reply]
"The cloud" is just a large space where the electron "probably or usually" is. All we can actually say is how likely it is for the electron to be a certain distance away, or else to say "the electron is unlikely to be [somewhere]" for some essentially arbitrary meaning of "unlikely" (there's not literally an "inner edge" of the cloud the way your question suggests). DMacks (talk) 21:13, 8 February 2013 (UTC)[reply]
There is no such thing as an "infinite" distance. But they might mean that there is no limit to how far the distance could be, which is not quite the same thing mathematically. ←Baseball BugsWhat's up, Doc?carrots→ 22:10, 8 February 2013 (UTC)[reply]
To put it in context, it's overwhelmingly likely that the electrons are closer to the nucleus than the nucleus of the next atom along - but there is a vanishingly small chance that the electron could be over on the other side of the universe. There is no particular paradox to this. The "distance between atoms" is likely to be measured from nucleus to nucleus. SteveBaker (talk) 00:08, 9 February 2013 (UTC)[reply]
Is "the" electron in a hydrogen atom really any random electron in existence and not necessarily the same over time?GeeBIGS (talk) 02:48, 9 February 2013 (UTC)[reply]
Why isn't empty space listed as a critical component of an atom? Since without it you can't have an atomGeeBIGS (talk) 02:54, 9 February 2013 (UTC)[reply]
It's not a component of an electron. Quarks could also be a component of everything that exists,right? but you don't just take them for granted or as you call it understood, what's the difference?GeeBIGS (talk) 07:08, 9 February 2013 (UTC)[reply]
IIRC, atoms and the like are pretty much localized to planets, etc.; most of the matter in "outer space" is more accurately Plasma_(physics)#Common_plasmas. (i.e., nuclei and electrons are just floating around promiscuously instead of in committed relationships) Gzuckier (talk) 04:39, 11 February 2013 (UTC)[reply]
DVI and VGA cable
Using a DVI cable instead of a VGA cable makes any difference in quality, performance or anything?
The Computer Ref Desk might be a better place to post this Q. StuRat (talk) 23:02, 8 February 2013 (UTC)[reply]
VGA is analog, DVI is digital. If you use VGA the monitor will convert it to digital internally, so it can't look any better than DVI. Whether it will look worse depends on the quality of the analog-to-digital converter (and how picky you are). I've seen wide variation in this, and I can't tell you anything about that particular monitor. As far as performance goes, it won't have any effect on frame rate. There could be a tiny difference in the lag time between the video card emitting the signal and the monitor displaying it, but I don't know which interface it would favor. -- BenRG (talk) 05:45, 9 February 2013 (UTC)[reply]
Use DVI if you can. If the source can produce high definition video it may only do that for DVI or the image may be slightly degraded comparatively even if it does send it over a VGA cable. Dmcq (talk) 11:56, 9 February 2013 (UTC)[reply]
Amount of heat input required to cut stainless steel plates
The first row of the table in this section seems highly counter-intuitive. If I'm reading it correctly, it seems to state that as plate thickness increases you actually need less laser power to cut through it. I've check the referenced source[13] and it is not a typo. Is this actually true? I'm having a hard time believing it despite the authoritative source. Is there an explanation for this? Dncsky (talk) 23:47, 8 February 2013 (UTC)[reply]
That does seem a little odd at first sight. I actually own a laser cutter (a 120Watt lasersaur)...and although I don't use it to cut metal - I think I may have some insight here.
I know that when cutting acrylic plastic with my laser cutter, the slot that the laser cuts acts like a waveguide - keeping the beam tightly in focus rather than spreading out. That means that you don't need much (if any) additional laser power to cut thick acrylic than thin...within reason.
When cutting wood and metal, it's common to blast a "cutting gas" into the slot where the laser is cutting - for cutting wood, you can use nitrogen to exclude oxygen from the cut and thereby prevent scorching. Or you if you're a cheapskate like me you can just blast air into the slot instead! A lot of people who are cutting metal use pure oxygen as their cutting gas. The gasses inside the cut become insanely hot and this "gas assist" lasering process pushes that hot gas into the path of the laser so that the material is pre-heated by the time the laser reaches it.
When you're cutting a thick, non-combustible material, the gas assist works much more effectively than for thin materials - and for shiney materials, the "waveguide" effect keeps the laser well focussed throughout the thickness - which is another big win. With metals, a consequence of that is that you need a heck of a lot of laser power to make that first hole (the "pierce") - but after that, you can dial it way back for the actual cutting. I'd be surprised if you could cut steel at all with a 250 watt laser without pre-heating it enough to get that first pierce hole and allow the waveguide and hot gas channelling to start working for you.
But here is the key: The business of delivering laser power onto the target is as much a matter of speed as power. My laser (which I predominantly use for cutting plywood) can cut effectively at a speed of 2,000 mm/sec only if the laser is operating at 100% full power. But you can also cut the same thickness of plywood at 50% power if you drop the feed rate to around 800 mm/sec. In effect, you're holding the laser in one place for longer so that the total energy delivered to the material is about the same. Balancing speed versus power is important for some materials that are liable to burn or melt or buckle when they get too hot.
So I suspect that although the laser power can be much smaller for thicker materials, the feed rate is probably a heck of a lot slower too.
In fact, that's backed up beautifully by our article - which shows 1,000 inches per minute as the feed rate for the thinnest steel and just 18 inches per minute for the thick stuff! In terms of Watt-seconds per linear inch of material cut (now that's a weird unit!), then using the data from the table, for the very thin steel, it's taking 1,000 Watts for 60 seconds (60,000Ws) to cut 1,000 inches of thin steel. That's 60Ws per inch of cutting. For the thickest steel, you need 250 Watts for a 60 seconds (15,000Ws) to cut for just 18" - which is 833Ws per inch. In other words, you need about 13 times as much laser energy (measured in Watt-seconds) to cut a 1" long slot in 0.25" steel than you do to cut the same slot in 0.02" steel...which (un-amazingly) is the same amount of energy per square inch of material removed - no matter the thickness.
The table is really saying that for very thick material, you can cut MUCH more slowly - and if you do that, you can get away with less laser energy. Doubtless you could cut 0.02" steel with a 250 Watt laser if you went slowly enough...and 0.25" steel would cut more quickly with a 1,000 Watt laser. But going so slowly with a thin sheet of material might produce other problems such as buckling and undesired melting away from the edge of the cut-line. So probably you have to use a lot of power to cut it quickly and thereby avoid those kinds of problem. Thicker material is unlikely to buckle - so you can take your time and use a smaller laser.
Perfect answer. Thanks a lot!Dncsky (talk) 01:27, 9 February 2013 (UTC)[reply]
It looks like your cutting gas has a couple of uses. The Wood/nitrogen case is clearly to stop combustion, but the steel/oxygen case turns the laser cutter into a Oxyfuel cutting torch. One of those can cut steel up to 2 feet (0.66 meters) thick because the steel itself works as a fuel. --Guy Macon (talk) 03:53, 9 February 2013 (UTC)[reply]
Yes, I believe so. When I was considering using Nitrogen as a cutting gas on my lowly 120 Watt beast, I wanted to work out how long a $100 cylinder of the stuff would last me (answer: Not long enough!) - and I got into an online discussion with a guy who was actually talking about cutting thick steel. He came up with numbers that implied to me that he was using a gas stream at close to the speed of sound - blasting down through the same hole that the laser beam emerges from - so it's a properly coaxial affair. Clearly that amount of pressure with pure oxygen and with all the heat that several kilowatts of laser energy adds will indeed produce something very similar to an oxyfuel system.
I soon realised that he and I were talking about entirely different systems! My machine produces a perfectly useful "air-assist" for cutting 3mm plywood using nothing more than a $77, 0.7 cubic feet per minute airbrush compressor - feeding a coaxial air stream through a 5mm hole...not much more than a gentle breeze! But even that relatively modest gas jet produces a dramatically better cutting speed than the laser alone (and incidentally blows smoke and debris away from the horribly expensive gold coated zinc-selenium lens). So there is quite a bit of subtlety to how these systems work.
Even on my lasersaur, the means by which the laser cuts varies from material to material. When cutting wood with an air jet, it combusts the material, producing smoke, water vapor, carbon dioxide, some kind of yellowish tar-like substance that coats my machine and has to be cleaned off regularly. It leaves behind a very smooth, but somewhat charred, edge. However, if you do use nitrogen to prevent combustion, the wood just "goes away" without producing smoke and the edge of the material is very clean. When cutting acrylic plastic, the laser produces some kind of chemical change that's not just combustion - and the result is a perfectly clean cut with shiny, smooth, clear edges. With some other plastics, it simply melts the material without any chemical change whatever. SteveBaker (talk) 16:32, 9 February 2013 (UTC)[reply]
Ack! Sorry - that should have been 2000mm per MINUTE not per second! But in any case, I'm not cutting metal - thin plywood, acrylic plastic, cloth, paper, cardboard, etc. You can see the stuff we make with it at http://renaissanceminiatures.comSteveBaker (talk) 04:58, 10 February 2013 (UTC)[reply]
February 9
Organs Questions
I've got three questions about organs and whatnot?
1. If someone old dies and then an organ/body part of his/hers gets put into another body, does this organ/body part still function based on its actual age or does being put in a new body makes it function as if it were of a younger age?
2. What is the age limit for donating blood, bone marrow, and various organs? Can a 100 year old's donated organ(s) or blood work successfully/well in another person's body?
3. Does any country maintain a list of the blood types and organ donation status of all or most of its population? I'm tempted to think No and that it would be considered a violation of privacy to do something like this, but I am still interested in finding this out for sure.
1. Yes to the former, age is not a state of mind, it is a biological fact, you can't rejuvenate organs in that way. Plasmic Physics (talk) 00:33, 9 February 2013 (UTC)[reply]
Thank you very much. Here's another question--do the various human organs age and decline in the same way that humans do? For instance, a healthy human who takes care of himself/herself would probably reach age 85 or 90, but afterwards it becomes a very uphill climb, with very few people who reach age 90 reaching age 100 and very few people who reach age 100 reaching age 110. Do most human organs (if put in a functioning body) also stop functioning when they (meaning the organs) are 85-110 years old? Futurist110 (talk) 00:42, 9 February 2013 (UTC)[reply]
I'm not an expert in this this field, but I would say yes, which is why younger transplants are prefered - they have a longer useful lifetime. Plasmic Physics (talk) 00:51, 9 February 2013 (UTC)[reply]
That's believable, because I know for certain that different organs age at a different rate. Take a car for example, the tires get worn out faster than the transmission (usually). Plasmic Physics (talk) 01:47, 9 February 2013 (UTC)[reply]
But this contradicts your previous statement of appearing to agree with me that human organs probably age as fast as human themselves do. That said, theoretically it might be possible for a human male to reach age 123, but this hasn't happened before since by that point it would be extremely rare for all of a human male's organs to still be working/alive. Futurist110 (talk) 02:10, 9 February 2013 (UTC)[reply]
I never stated that they age at the same rate, only that they age. Otherwise, please name the time stamp of my comment which is contrary. Plasmic Physics (talk) 05:59, 10 February 2013 (UTC)[reply]
In the UK the age limit for blood donation used to be 70, but they scrapped that (ref), and it's now based on the donor's health. Crucially, certain medications for chronic conditions (medications that are very commonly taken by people as they age) disqualify people as donors, so in practice the number of donors older than 70 is pretty low. -- Finlay McWalterჷTalk 00:38, 9 February 2013 (UTC)[reply]
So in the U.K. someone aged 100 or above but in good health can (sometimes) donate his/her blood? What about organ donations and bone marrow donations? Futurist110 (talk) 00:42, 9 February 2013 (UTC)[reply]
As I thought I had remembered, and it's confirmed in Red blood cell, blood is constantly being created and recycled, so it might be that any donation age limits are concerns about the general well-being of the donor, rather than any concerns about "old" blood. ←Baseball BugsWhat's up, Doc?carrots→ 06:13, 9 February 2013 (UTC)[reply]
When DFDBA (demineralized freeze-dried bone allograft) is procured from cadavers, a molecule known as BMP (bone morphogenetic protein) acts to trigger greater osteblastic potential in the recipient site, theoretically leading to greater bone growth following grafting. The BMPs are even able to trigger de novo bone growth even in the absence of osteblasts by stimulating regular fibroblasts to transform, and studies have shown ossicle formation in murine muscle pouches (in the absence of bone tissue). Anyway, it was shown by Schwartz in 1996 (J Perio) that the age of the cadaver donor correlated with the osteoinductive potential of the bone graft because of the relative potency of the BMPs. He studied the results of DFDBA from both genders and various ages and determined that the dentist (or other clinician) will not know the effects because he or she won't be able to pick the age of the donor. DRosenbach(Talk | Contribs) 15:11, 10 February 2013 (UTC)[reply]
Particle-wave duality
In an atom, it is said that the electron has a probability of being located at a position. Is that technically, or heuristically, correct - is the electron truely whizzing about the atom in a random way, or is it everywhere around the atom at once? I just find it hard to imagine that the electron retains its particle identity. I consider an unbound electron to be like an icy small solar system body, and a bound electron to be like the same SSSB vaporised and converted to the atmosphere of a planet. So, does the probability indicate how often the electron is expected to be encountered at that location, or how much of the electron is to be encountered at that location? Plasmic Physics (talk) 00:30, 9 February 2013 (UTC)[reply]
It's dangerous to think in those "normal" terms. It's not particularly meaningful to ask where the electron "is" or how much of it is where - because it's really only a probability field - and it (in a sense) "teleports" between locations within the probability cloud. For example, there is an effect called "Quantum tunnelling" (which is what makes the flash memory in your phone/tablet/memory-stick work - so we know it's true!). In very simple terms, the flash memory cell has a barrier between two locations that (classically) the electron cannot cross - but it can teleport between those two locations if properly coerced because there is always a finite probability of it being on the other side of the uncrossable barrier! So this is a very real effect - large, real-world things like telephones rely utterly on this weird quantum behavior. But it's quite hard to reconcile with day to day life...and that's the core problem with understanding quantum theory. At those levels of existence, things are very, very weird - and trying to get your head around a physical understanding of it is impossible. The only real way to get a handle on it is via the math. SteveBaker (talk) 01:02, 9 February 2013 (UTC)[reply]
So, a bound electron has a precise location that changes with time? I know about quantum teleportation, I just never reconcilled it with the behaviour of bound electrons. They never discussed electron motion within atoms in the lectures. Plasmic Physics (talk) 01:42, 9 February 2013 (UTC)[reply]
Quick aside — what's being discussed is quantum tunneling, not quantum teleportation, which I think is some information-theoretic abstraction rather than an electron actually showing up in a different place.
Whether an electron has a precise location is a subtle question that gets into interpretations of quantum mechanics. I think most physicists generally prefer to avoid the question altogether ("shut up and calculate" says the shade of Feynman) and leave it to the philosophers. --Trovatore (talk) 03:22, 9 February 2013 (UTC)[reply]
Is there a way of testing either alternative? Plasmic Physics (talk) 03:48, 9 February 2013 (UTC)[reply]
There are more than two alternatives. As far as I know, no experiments have been devised that would distinguish among any of them. Strict positivists probably consider it a meaningless question for that reason, but for people with a more realistic view this is hard to swallow. But it's hard to get realism to play nice with QM under any conditions. --Trovatore (talk) 04:03, 9 February 2013 (UTC)[reply]
I should say, though, that your notion of a sort of "electron vapor", spread out over space, is not really one of the alternatives. Or at least I don't think it is. The electron itself is very tiny — as far as anyone knows, pointlike. The probability distribution tells you how likely it is that the electron will be found at a given location, but it is not the electron itself. --Trovatore (talk) 05:12, 9 February 2013 (UTC)[reply]
The thing is, the "wave" vs. "particle" thing is a false dichotomy. There are not "alternatives" and these are not the two. Electrons always behave like electrons. The problem is that the way they behave simply does not have analogues you can experience with your 5 senses. That is, there is nothing in anything you have experienced, and thus nothing you can "visualize" which fully captures what an electron is. The best we can say is that, in some applications it is helpful to visualize the electron as a particle (but only for those applications) and for other applications, it is helpful to visualize the electron as a wave. Those are still human-created models, however. The electron doesn't change its behavior, it doesn't swap "modes" or jump between "alternatives": it just goes on being an electron doing what electrons always do. It's our problem to come up with models to explain it. And it's been about a century since physicists stopped trying to create a "picture" of what an electron is. That's the whole spirit of the "shut up and calculate" exhortation is above. Electrons are best modeled by mathematical equations that describe and predict what they do: that there is no single nice visualization you can make which captures the full nature of "electronness" is ultimately not the electron's problem. --Jayron32 05:22, 10 February 2013 (UTC)[reply]
The picture of the electron in a hydrogen atom as a stationary wave and the picture of it as a particle randomly moving around are about equally valid. They can't be tested against each other—they're just two different ways of looking at the same physics.
Quantum mechanics actually has two different classical limits. If you take ħ → 0 while keeping the E in E=hf constant (so that f → ∞) you get the classical particle limit: each particle still carries energy E, but the infinite wavelength means you can never observe any wave interference behavior. If you take ħ → 0 while keeping the f in E=hf constant (so that E → 0) you get the classical wave limit: there are infinitely many particles each carrying infinitesimal energy that smoothly cover everything so you never see shot noise. Quite a lot of behavior that's often described as "purely quantum" is really behavior that vanishes in the particle limit but survives in the wave limit. Tunneling is one example of that: in the classical context it's called Evanescent-wave coupling. Of course there are other things that vanish in the wave limit but survive in the particle limit, like shot noise. Quantum mechanics sort of sits halfway between these two types of classical theories, so it's helpful to keep both pictures in mind. -- BenRG (talk) 05:15, 9 February 2013 (UTC)[reply]
I'm worried that this answer sounds overly compatible with the "electron vapor" idea. A stationary wave is not the same as a spread-out electron, any more than a radio wave is a spread-out photon. --Trovatore (talk) 05:26, 9 February 2013 (UTC)[reply]
A radio wave containing one quantum of energy is a spread out photon. A radio wave containing many times that much energy is a bunch of spread out photons. Photons are not localized in a coherent wave.
I'm not sure exactly what you mean by the electron vapor idea but I think there's nothing especially wrong with that idea. Feynman famously said that the single most informative statement about the world is "all things are made of atoms — little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another". That's a picture of atoms as somewhat sticky, somewhat rubbery balls. Sure they're quantum objects but that doesn't mean these aren't useful classical analogues for various aspects of their behavior. -- BenRG (talk) 06:56, 9 February 2013 (UTC)[reply]
Well, no, come on, I know you're a physicist and I'm not, but a radio wave is not a spread-out photon. A radio wave is electric and magnetic fields. An electron "wave" is an area of probability density. It's not a particle spread out. That's a cheap way of restoring realism that doesn't work.
You can restore realism with the many-worlds interpretation, by letting the real thing be the entire collection of worlds rather than just ours. Maybe you can restore it by the transactional interpretation; I've never understood it well enough to be sure. But not with the image of a "spread-out particle". It's just the wrong image; it doesn't lead to intuitions that correspond with theory or experiment. --Trovatore (talk) 09:12, 9 February 2013 (UTC)[reply]
I'm not a physicist, I just studied it in college. First, as I mentioned in another thread, the Standard Model is a quantized classical field theory. The wave nature of the electromagnetic and electron fields (the latter has no good name) is already there at the classical level. Quantization adds particle-like behavior, but the wavelike behavior doesn't disappear. You can and should think of a photon as the dimmest possible electromagnetic field. It's a misleading name for it, but it's too late to change that. If you want to talk about the pointlike behavior of light (i.e., shot noise), it would probably be better to talk about the "darkenon", which is one quantum of energy transferred from the electromagnetic field to a single silver halide crystal. (I just made that word up.)
Quantization is always the same. You probably wouldn't say that phonons are points and the bulk vibration of the crystal is just an indication of the probability of finding one in a particular place, but it's just as true for them as it is for electrons or photons. Or quantum vortices, which don't seem to have a commonly used name ending in -on but are particles nonetheless. -- BenRG (talk) 18:32, 9 February 2013 (UTC)[reply]
A radio wave totally is just a spread-out photon (or, more commonly, for normal amplitudes, it is a superposition of many photons). It's rarely useful to analyze this way; and we rarely have a wave at RF frequencies whose amplitude is so low as to be a single quantum emission; but this is still definitionally true and is easier to build a single RF photon emitter, in practice, than a single-photon-emitter at optical frequencies. Any electromagnetic wave can be viewed as a wave or as a particle. Considering a macroscopic wavelength as a "particle" can be very instructive in developing intuition about wave-particle duality. BenRG's comments about phonons are spot-on. Nimur (talk) 20:51, 9 February 2013 (UTC)[reply]
I disagree. It's a bad guide to intuition. It avoids confronting the failure of local realism. It sounds as though there's all this real "stuff" that's spread out over space, and that just isn't true. Also it completely clouds (ha!) the issue when you try to extend it to non-pointlike particles, like the proton — the learner can be forgiven for saying, wait a minute, here you're telling me about the proton "radius" (as distinct from the electron radius of, presumably, zero), but at the same time you have a picture of the proton as spread out over space, are these the same thing or not?
Particles are ipso facto localized, but the place they're localized to is, depending on interpretation, uncertain, different in different "worlds", or something else. --Trovatore (talk) 21:00, 9 February 2013 (UTC)[reply]
Well, I mean, the world is as real as it ever was, and failing to acknowledge that seems like an error to me. Of course there's stuff out there; it's what we've been trying to understand with all this physics nonsense. Saying that we've discovered by studying it that it isn't real, aside from being philosophically silly, doesn't tell you anything. It's wakalixes. What Feynman said about atoms does tell you things. The stickiness of macroscopic objects derives from the stickiness of atoms. Solids tend not to be sticky because they have rough surfaces and there's very little atom-to-atom contact when you press them together. Liquids stick to solids because they flow into the gaps. Gas molecules don't stick to each other because they're moving too fast, but if they slow down enough they do stick together, and that's why gases liquefy when you cool them down. And so on. Atoms are solid objects. Macroscopic solids are made of atoms and derive their solidity from the solidity of atoms (which is due to the Pauli exclusion principle). It's ridiculous to deny all of this or to imply that there's some kind of connection to many-worlds or what have you. We've actually learned some things about the everyday world in the last hundred years.
It's true that protons have an intrinsic (interaction) radius and also are spread out and that those things are independent. Saying that you shouldn't tell people about that because it's confusing isn't a very good argument. Mathematically the interaction radius shows up in the interaction part of the Lagrangian as a coupling of field values at different points in space. The photon-electron interaction in QED is a product of field values separately at each space(time) point, integrated over all the points, and that's the sense in which they are pointlike. A free field/particle is not pointlike in any way. That's why I said above that it's probably better to treat the interactions as pointlike rather than the particles themselves.
In many ways Standard Model particles/fields behave like classical particles. But saying that the particle picture is more fundamental, and the wave picture secondary, is completely wrong. -- BenRG (talk) 22:17, 9 February 2013 (UTC)[reply]
Wait a minute, I didn't say there wasn't real stuff out there. I said a delocalized pointlike particle is not "real stuff spread out". I stand by that. A point particle is a point, no matter how spread out its probability density function is. --Trovatore (talk) 22:27, 9 February 2013 (UTC)[reply]
Okay, look. What you seem to be saying is that fundamental particles don't occupy space—which means that the term "occupying space" is useless, since nothing does it—but that they do something, presumably involving the Pauli exclusion principle, that creates the impression of occupying space. I suggest calling that thing that they do "occupying space". I don't see what other meaning the term can have.
As far as I can tell, you haven't explained why you think the particles are just points even in the face of all the arguments to the contrary that I've presented. If it's because you think the wave function is a probability density, I think you're forgetting that it's a probability distribution over classical field configurations, not classical particle coordinates, since the Standard Model is a quantized field theory. If it's because of Feynman diagrams, I'll explain the reasons why Feynman diagrams are unlikely to be fundamental. -- BenRG (talk) 00:04, 10 February 2013 (UTC)[reply]
Well yeah, the path integral formulation makes a lot of sense to me. Why do you see that as non-fundamental?
But you know, anyway, your point about a probability distriubtion over field configurations — that's just a different basis for the vector space. Surely the appropriate basis when you're calling them particles, specifically, is the one where the position distribution is a delta function. That's kind of what "particle" means. --Trovatore (talk) 01:36, 10 February 2013 (UTC)[reply]
To particle physicists "particle" and "field" are nearly synonymous. You're taking the terminology too literally. Atoms aren't indivisible, protons aren't fundamental and particles aren't points. Also, the wave function has an independent value at all field configurations, not just a set of basis states. There's no basis for the quantum Hilbert space that limits you to delta functions in physical space.
Individual Feynman diagrams break gauge symmetry and in some cases involve fictitious particles. The symmetry is restored and the fake particles disappear in the sum, but it's not very plausible that the universe is really adding a bunch of asymmetric diagrams with extra particles and ending up with a perfect symmetry. It's similar to treating the expansion as fundamental. Very similar, in fact, because Feynman diagrams label terms in a series expansion of the path integral. The basic idea is easy to understand. Start with . If H is not a function of time, the solutions are . You can expand that as , i.e., as a sum of ψ0, ψ0 acted on once by −iHt, ψ0 acted on twice by −iHt, .... If H = H1 + H2 then the sum includes ψ0 acted on once by −iH1t, once by −iH2t, twice by −iH1t, once by each, etc. This is very sloppy since I should be using the Lagrangian and I should separate the interacting and noninteracting portions, but I hope you see that in a series expansion you quite naturally get a sum of all finite combinations of discrete interaction terms. That's not to say that the boson-exchange picture suggested by the simplest diagrams is wrong. At least in QED scattering it's correct inasmuch as it's a good approximation, but it's not The Truth any more than E = ½mv².
Another problem is that there are phenomena in the Standard Model that don't have an expansion in Feynman diagrams, such as the Higgs mechanism. (Which is not the same as the Higgs particle, but I think the detection of a particle so similar to the predicted one proves the correctness of the Higgs mechanism beyond a reasonable doubt.)
'Electron vapour' is exactly the name I would give it - a bound electron completely evaporates, from a particle into a vapour, surrounding the nucleus where the cloud density is equal to the 'probability'. Plasmic Physics (talk) 09:18, 9 February 2013 (UTC)[reply]
But that's exactly the wrong image. If you use that image, your intuition will lead you to incorrect conclusions. The electron has (as far as anyone knows) zero volume — it's just a point. That's true whether it's bound or free.
The "cloud" is a quantum superposition of places the electron might be. But each of those places is just a point (again, as far as anyone knows). --Trovatore (talk) 09:24, 9 February 2013 (UTC)[reply]
Yes, that's wrong—the electron's nature doesn't change just because it's part of an atom. If it's a cloud in that situation, it's a cloud the rest of the time too. If it's a point particle the rest of the time, it's a point particle when part of an atom. -- BenRG (talk) 18:32, 9 February 2013 (UTC)[reply]
There are several mistakes here. You're forgetting the solid-illusion, nothing is truely solid, especially electrons. Electrons consists of various fields that are concentrated in a very small space without definite boundaries. The 'radius' which comes in at a maximum of 10-22 m, is completely arbitrary, and was chosen to represent a distance from the centre of the fields to where the electric field decayed sufficiently. Ergo, the electron itself is a tiny version of the 'electron vapour'. The only difference between the bound and unbound electron would be the diffusivity gradient and shape depending on which orbital was occupied.
P.S. From the article, 'point particle' is just a heuristic to describe the relative dimensions of a partcile. Plasmic Physics (talk) 19:36, 9 February 2013 (UTC)[reply]
lemme toss the uncertainty principle into the mix; the uncertainty referred to is split between the position and momentum of the electron (in this instance). So, if you know the position of the electron to a high degree of accuracy, you have no idea what the momentum might be (i.e., is it moving or not?); alternately, if you know the velocity of the electron, you can't know what the position is, i.e. is the position changing. So, "is the electron truely whizzing about the atom in a random way, or is it everywhere around the atom at once"? You're asking for both position and momentum simultaneously, which makes this a question which cannot be answered, or more accurately, a question which is basically meaningless at this scale. Gzuckier (talk) 04:59, 11 February 2013 (UTC)[reply]
Covering bare floorboards
Okay, so basically today I put the wrong kind of soap in the dishwasher (liquid instead of powder :( ) and apparently the dishwasher didn't like that. I found this out as an hour after I started the load, I got up to check it, and found the kitchen floor F**KING FLOODED IN LATHER AND SUDS! Needless to say, i put away the dishes and cleaned up the mess, and the kitchen floor was so clean afterwards I could see my reflection in it. It was then I decided that as long as I was working around the bottom of the dishwasher, I might as well nail down the curled up linoleum around the edges of the floor. So I got out a box of 1&1/4" nails and a hammer and started pounding away.
However, some portions of the floor around the dishwasher and sink were so curled, distorted, and cracked that I couldn't nail them down, I had to cut off and throw away the ends before doing so. This led to a small, but long area of bare floorboard under the sink and dishwasher. This is a problem because:
The floor is particle board, so can cause splinters in the feet if stepped on wrong.
In addition to the water damage that the floor has doubtless suffered through the years already, it is now completely bare and any liquid spilled on the floor in that area could damage it.
So anyway, my problem is I can't decide which substance would be best for a sort of footstop under the counter. What I want to do is take one of those inclined thresholds like the sort under a front door to a house and nail it down there. I need a footstop which can prevent water damage and accidental kicking of the baseboards, hold the linoleum down indefinitely without cracking it, and last for up to five years. My options are:
Metal
Rubber
Plastic
Treated Wood
More linoleum flooring
Tile
So, which would be best? (PS I filed this under science because I thought this was an engineering topic) --FreeWales Now!what did I screw up? 00:56, 9 February 2013 (UTC)[reply]
Free Wales now? Is Jimbo in jail? Oh, wait... :)
Will a high dam sill[14][15][16] do what you want to do? They come in long versions for use with garage doors. --Guy Macon (talk) 02:30, 9 February 2013 (UTC)[reply]
A bare margin with some sort of sill is not going to do you well if you have another flood. You're probably better of with a new floor that sits flush against the counter although I am sure you don't want to hear that. I won't laugh at your situation having gone through it myself and had a friend go through it last month. μηδείς (talk) 02:49, 9 February 2013 (UTC)[reply]
Thank you, that high dam sill thing does sound exactly what i'm looking for. Since my dishwasher seems to have flooded from under the door rather than the baseplate, needing a new floor shouldn't be a problem. Pretty much all I needed was a durable "ramp" to allow water to drain out onto the rest of the floor and off the baseboards while looking decent. Heading to Home Depot tomorrow... --FreeWales Now!what did I screw up? 03:41, 9 February 2013 (UTC)[reply]
It is altogether possible that some hose behind will develop a small leak at some time and what you are doing would hide the damage and make a pool to soak into the floor. Just fix the floor properly and cover the edge between the wall and the floor behind to stop any small leak into the crack. Dmcq (talk) 11:38, 9 February 2013 (UTC)[reply]
Note that the Refdesk does not give advice, accepts no liability if black mold spreads throughout your house causing serious and permanent illness that can be palliated only by expensive visits with a mind-body therapist, and we are not licensed interior designers[17] and therefore are not allowed to give interior design advice in Florida. Except in residences. Or is it Virginia now? Heck, it's a racket every bit as legitimate as the "medical advice" thing they're always on about here. Wnt (talk) 17:35, 9 February 2013 (UTC)[reply]
why was rutherford model of an atom discarded ?
Rutherford model was discarded because according to that model electrons will radiate energy as they move around the nucleus. In Bohr model, electrons revolve around the nucleus without radiating any energy. How is this possible ? The case (radiating energy), which was applicable for Rutherford model, was not applicable for Bohr model, why. --Concepts of Physics (talk) 03:05, 9 February 2013 (UTC)[reply]
For convenience, see Rutherford model and Bohr model. Basically, the Bohr model was just more specific than the Rutherford model. Rather than predicting the electrons existed in a cloud, possibly orbiting the nucleus, Bohr predicted very specific orbits. His model never explained how stably orbiting electrons fail to radiate energy, which was part of why physicists knew there was still a deficit in their understanding of the electron. Someguy1221 (talk) 03:18, 9 February 2013 (UTC)[reply]
An easy way to think of the historical chronology of these models is this:
Thomson: "atoms are not indivisible, but are made up of negative subatomic particles in complicated orbits, all within a diffuse positive field"
Rutherford: "actually, atoms are a dense positive nucleus around which tiny negatively subatomic electrons orbit in a classical way, but I don't know why they don't lose energy"
Bohr: "ah, they don't lose energy because they're quantum, not classical, and in fact the way they gain/lose energy has to do with discrete stable states. But I don't know why that is other than to say that the quantum world is strange."
De Broglie: "ah, the reason they are distinct stable states is because there are only so many stable wave functions, and the electrons are basically waves"
It's not so much that Rutherford was "discarded" so much as "built-upon." Ditto with Bohr's, which wasn't the last step there either. (Nor was de Broglie's, of course.) Obviously I'm simplifying the technical stuff here but such is how it goes with simplifications! --Mr.98 (talk) 18:05, 9 February 2013 (UTC)[reply]
In fact, the "plum pudding" model works pretty well for many applications in classical mechanics. You only need to worry about the breakdown of this model when you perform experiments with resolution better than the scale length of the atom. That means you're sending down photons of particular wavelengths (so you need the mechanics of Compton scattering; or you're analyzing emission spectra of atoms with high resolution. If you apply classical theory to the plum pudding model, and your scale length is always sufficiently larger than the atom, you will never find any conundrums! In formulations of quantum mechanics, this is an important and often-underemphasized empirical fact. Mathematically, this corresponds to a boundary condition, or a limit case, of any equation proposed to describe a quantum mechanical system: when extrapolated to large numbers, or large sizes, the formulation must remain consistent with what we observe! It is my opinion that most of the "conundrums" people encounter when they try to study quantum mechanics would be straightforwardly resolved if people would review the mathematical conceptual leap in the limit formulation of these problems. Nimur (talk) 20:01, 9 February 2013 (UTC)[reply]
Quantum theory revolutionized physics at the beginning of the 20th century, when Max Planck and Albert Einstein postulated that light energy is emitted or absorbed in discrete amounts known as quanta (singular, quantum). In 1913, Niels Bohr incorporated this idea into his Bohr model of the atom, in which an electron could only orbit the nucleus in particular circular orbits with fixed angular momentum and energy, its distance from the nucleus (i.e., their radii) being proportional to its energy. Under this model an electron could not spiral into the nucleus because it could not lose energy in a continuous manner; instead, it could only make instantaneous "quantum leaps" between the fixed energy levels. When this occurred, light was emitted or absorbed at a frequency proportional to the change in energy (hence the absorption and emission of light in discrete spectra).
Bohr's model was not perfect. It could only predict the spectral lines of hydrogen; it couldn't predict those of multielectron atoms. Worse still, as spectrographic technology improved, additional spectral lines in hydrogen were observed which Bohr's model couldn't explain. In 1916, Arnold Sommerfeld added elliptical orbits to the Bohr model to explain the extra emission lines, but this made the model very difficult to use, and it still couldn't explain more complex atoms. Want to be Einstein (talk) 10:09, 10 February 2013 (UTC)[reply]
This reminds me of my favorite piece of dumb management/marketing speak, where something or other is described as a "quantum leap" in blahblahblah. Someday I'll stick up my hand and say "You mean it's the smallest possible change?" Gzuckier (talk) 05:05, 11 February 2013 (UTC)[reply]
Diameter is not a determinant for destructiveness against armor; note that sabot anti-tank rounds have a "shoe" (fr. sabot) that separates from the round to release a kinetic energy projectile which is smaller than the diameter of the barrel from which it is fired. From the article: For reasons why a smaller diameter projectile can be desirable, see external ballistics and terminal ballistics. ~E:74.60.29.141 (talk) 06:02, 9 February 2013 (UTC)[reply]
It probably wouldn't be sufficiently effective against a modern MBT. This 1980 report about an attack by A-10s on Korean War era M47 Pattons says the 140 hits, only 17 penetrated the Patton's armour. A Patton has 100mm of steel armour; the later M1 Abrams variants have advanced reactive Chobham composite armour which give a protection equivalent to 600-1000mm of steel armour. That's why the A-10 carried higher performance weapons like AGM-65 Maverick as well, specifically for hard targets that the GAU-8 wouldn't kill. The GAU-8 is still useful against a wide variety of battlefield targets like AFVs, APCs, trucks, tankers, jeeps, artillery pieces, tractors, and self-propelled guns, and against unfortified structures. -- Finlay McWalterჷTalk 16:33, 9 February 2013 (UTC)[reply]
That said, it's not totally useless against an MBT; given the GAU's weight of fire, it's quite possible it would be able to detrack an MBT, rendering it ineffective and vulnerable. -- Finlay McWalterჷTalk 16:36, 9 February 2013 (UTC)[reply]
The M829's KE penetrator is only 20 mm in diameter. However the A-10's cannon is ineffective at killing modern MBTs. It can handle out of date T-62s and the likes, but not a post-1980 MBT. It can several damage them, destroying their tracks, vision equipment, even weapons and engine (weakly armored grille). A machine gun can be completely destroyed by a hit and I don't think a crew would risk firing a big gun that has been damaged by an autocannon, the shell might explode before leaving the barrel. All of this is repeairable though, for an assured detruction it must used its missiles. You must remember though that MBTs are only a small percentage of the vehicles in an army. Lightly armoured and soft skinned vehicles are the bulk, and the cannon can effectively destroy these without wasting a missile.--Whichwayto (talk) 17:11, 9 February 2013 (UTC)[reply]
For a kinetic kill system like a DU anti-armour round, the crucial factor is the projectile's kinetic energy. If I'm doing the maths right, the KE (at the muzzle) for a GAU-8 round is about 0.2 MJ; for an M829A3 from an Abrams its about 12 MJ. -- Finlay McWalterჷTalk 17:27, 9 February 2013 (UTC)[reply]
Actually, in the case of the Hog's GAU-8, the most crucial factor is hitting the top armor rather than the frontal armor -- a tank's armor is weakest on top. 24.23.196.85 (talk) 20:22, 9 February 2013 (UTC)[reply]
ineffective at killing modern MBTs is incompatible with what follows that claim. A blind, motionless, gunless MTB might as well be dead. It's certainly a mission-kill, and it's an easy target for infantry with any kind of anti-armour. --Stephan Schulz (talk) 21:18, 9 February 2013 (UTC)[reply]
But, as he said, almost all of that damage is repairable by the tank crew. Just bolt on the spare machine guns, realign the main gun, fix the tracks, and you're ready to go. Even if some of it isn't repairable, the tank can still be useful; for instance, if the periscope is wrecked, the commander can still effectively direct fire by standing up in the cupola (good tank commanders do this anyways), and even if the tank is rendered completely motionless, it is still quite useful as a fully traversable, armored artillery piece. Whoop whoop pull upBitching Betty | Averted crashes 17:16, 10 February 2013 (UTC)[reply]
did Curiosity or any other mars probe have a few grams to spare out of their multiton payload on a microphone, so we could hear what Mars sounds like? 178.48.114.143 (talk) 06:46, 9 February 2013 (UTC)[reply]
Considering the rover is two tons, the mic would have added precisely two grams and a single analog input, and was not mission-critical, would not need to be pointed anywhere and can literally be in any crevice, does not need to even be automatically collected or read or affect any other mission requirements, can break or fail to function without consequences, yet if it happened to still work, would bring huge outsider interest in mars, I hope you will agree that I am simply smarter than NASA. That video, until the rickroll, was absolutely FASCINATING. I was like, wow, wow, wow. Two grams. Fuck you, stilted engineers. You didn't even try! 178.48.114.143 (talk) 09:01, 9 February 2013 (UTC)[reply]
I'm sorry to disappoint your hope, but I don't agree. I suspect that the predominant sounds would be those of the motion of the rover. Are those of interest? I'd be surprised if a Mars microphone picked up any interesting sounds. (dinosaurs? little green men chatting? ) Dbfirs 09:30, 9 February 2013 (UTC)[reply]
It's impossible to predict what scientific discoveries will be made in advance, or else they wouldn't be discoveries. For example, the LHC was predicted to discover the Higgs boson, a particle that's been part of the Standard Model for decades, and that nobody seriously doubted the existence of. If that's all it discovers, the LHC is a huge waste of money, but physicists are hoping for something unexpected that will revolutionize physics. What will that be? Well, if anybody knew, it wouldn't be a discovery. --140.180.247.198 (talk) 17:41, 9 February 2013 (UTC)[reply]
Mars... I can't believe I'm back on Mars. Three times before, this place almost killed me. I swore I'd never give it another chance to finish the job. Humns got no business being here. No business at all. — Michael Garibaldi. How different is that from Saigon...shit? --Trovatore (talk) 09:36, 9 February 2013 (UTC)[reply]
Yes, the sounds of the motion of the rover would have been very interesting. Out of curiosity, why do you think we wouldn't have heard wind, as in the video? Or explosions in the distance from volcanos nearby, which the rover cannot see visually? Acoustics is so so cheap. If nothing else, the sound of the gravel/sand/whatever under the rover's feet... This article says it is "probably still volcanically active today": http://en.wikipedia.org/wiki/Volcanology_of_Mars - so, out of curiosity, if the rover happened land within earshot of a volcano, you don't think having a mic onboard would be something anyone would be interested in hearing? No mic is simply irresponsible. It's not like I'm asking you to include speakers, so we can blast some music and hear its echoes if we are near a wall. I'm asking for a two gram mic. There is no excuse. Aren't there dust storms, too? You don't want to know what they sound like? No imagination around here. 178.48.114.143 (talk) 10:14, 9 February 2013 (UTC)[reply]
“Curiouser and curiouser!” Cried Alice: I find usage of the clichéout of curiositycurious interesting in a thread discussing NASA's Curiosity rover --Senra (talk) 22:14, 9 February 2013 (UTC) [reply]
If we prorate the ENTIRE budget of $2.5 billion just into the 2000 pound payload - which is a huge overstatement, imagine if a scenario where the actual final payload is just 100 grams. Making it 101 grams by adding a gram to what we consider, wouldn't have cost a further marginal $25 million or increased the budget by a full percentage point! Most of the budget is in developing the rover, delivery system, etc. So this is a huge, huge, huge, huge, HUGE overestimation. Still, by this HUGE overestimation, another 2 grams added to the payload would have increased the size of the payload by 1/500000th, or the prorated budget from $2.5 billion by another $5,000. A handful of people, hell, a single individual, would have gladly paid that amount to hear Mars. If the only place a single recording ever appeared was youtube, the advertising revenue would be more than $5,000 from it. Think about it!!! There is just no excuse, except ignorance and lack of creativity. 178.48.114.143 (talk) 10:30, 9 February 2013 (UTC)[reply]
"ignorance and lack of creativity", eh? And yet ironically they're the ones driving a robot on Mars and you're the one wasting your time ranting on an internet forum which isn't actually a forum at all.Dncsky (talk) 13:36, 9 February 2013 (UTC)[reply]
I wouldn't be so quick to discount the OP's opinions, though some of them could have been expressed in a more rational fashion. I, for one, wouldn't mind listening to dust storm or similar event taking place on Mars (imagine all the relaxation CDs off-brand record labels could sell in five-and-dime shops!), and I'm certain the cost would have been neglibible compared to the other components of the rover. Let's also not forget that technical know-how does not necessarily correlate with creativity or a propensity for brilliant ideas (for example, this is a product that smart people allow to exist). Evanh2008(talk|contribs) 13:58, 9 February 2013 (UTC)[reply]
This question has been asked before, here. Here's one of the responses:
"Plenty of people have pushed for including a microphone on a Mars mission, most notably the Planetary Society, whose microphone actually flew on the failed Mars Polar Lander. The Phoenix lander also included a microphone, but only because its descent imager (MARDI) happened to have a microphone on its circuit board. MARDI was turned off because it had a risk of interfering with IMU measurements during landing, which could have been fatal to the spacecraft. According to an interview I heard with a Planetary Society member (Emily Lakdawalla), the Society lobbied to have their microphone included on Curiosity, but it was rejected because the rover was already complex enough (she didn't elaborate on this, and I haven't been able to find a more detailed explanation). Personally I think that's a strange reason, because a microphone is one of the smallest and simplest electronic devices possible (just look at your cellphone to see how small it could be). Having one would be great for PR, and NASA only gets funding if the public is excited and inspired by its missions."
Here's an answer from the Curiosity team themselves: [18]. The second part is interesting:
"Here's a little more info on the Phoenix microphone. It was essentially a hitch-hiker. It was built into another instrument taken off the shelf for the the lander, but it was never intended for the mission. There was no science team or budget connected to it. Since it was not intended for use it was never tested before launch and never entered into the power budget for the lander. Only after Phoenix successfully completed it's mission, 5 months after landing in the polar region, was the mission somewhat willing to test it. They couldn't do it earlier because they couldn't risk the prime goals of the mission if anything went wrong. The project manager was fairly certain it wouldn't work and was against trying it because he didn't want to raise expectations. His mind changed when we got a tweet to the @MarsPhoenix account from a man who said he was blind and how much he wished he could hear Mars because he couldn't see the pictures. A couple days later, they sent the signal to Phoenix to turn it on but we got.. well.. nothing. Empty files. If we had received anything, it would have been released. The team figured the mic was frozen solid and decided to give it a second try by leaving it on longer to warm up. Unfortunately, the Phoenix mission lost its last bit of power (as expected) before it got the second instruction." --140.180.247.198 (talk) 17:41, 9 February 2013 (UTC)[reply]
Let's rephrase the answer another way. Spaceship engineers are not stupid. They know what their instruments measure. Now, let's review this carefully, because it is not clear whether the OP knows what a microphone measures.
"Sound, obviously!" says the enthusiastic OP! "I want to record sounds! Put a microphone on the spaceship!"
Now, as the reference desk regulars who follow Science know, we've had a recent spate of questions asking about minimum densities that sound can propagate in. And, as we have repeatedly answered, there is no well-defined answer to that question. So, "is there sound on the Moon? Is there sound on Mars?" While our knee-jerk response is "of course not, these places are nearly total vacuums, and sound cannot propagate!" ... if we stop to really really carefully answer the question, we find that we must refine this to say, "...well, actually, the amplitude of acoustic waves in very-very-sparse atomspheres are very tiny, and the frequencies are very low."
This brings us back to my first point: spaceship engineers are very smart. So while you think "a microphone measures sound" and cuss out the engineers, let me defend their line of reasoning. A microphone doesn't measure sound at all. Most modern "2-gram" microphones are piezoelectric crystal attached to high gain differential amplifiers, and then digitally sampled. They measure a voltage, proportional to the mechanical stress on a piezoelectric crystal. When a sound is audible to a human, that sound corresponds to a pressure wave propagating in air, which will exert a mechanical stress on a collector plate and deform the piezo, inducing a voltage. That voltage will be amplified and sampled.
We can calculate exactly what voltage will be produced for any given deflection.
We know the density of the Mars atmosphere, and we know its pressure, and we know how much force a molecule can exert on a piezoacoustic coupling... and we know the electronic noise floor of the finest differential amplifier circuit this side of the Jovian moons... and we know the realistic sampling rate and quantization error for digitizers, because we studied elementary digital design theory... and the net result is: the signal will be all zeros.
If we know (from other experiments, and from application of the laws of fundamental physics, that it is not physically possible to measure signal, given the sparse atmosphere of Mars, we don't waste time putting a microphone on the spaceship.
Instead, we go back to basics: what are we trying to measure? Vibrations? Particle flux? We have instruments on Mars Science Laboratory to do that.MEDLI, RAD, and others.
So, the real point here is, the OP is making an unreasonable demand to place an instrument known not to work on a billion-dollar spacecraft. When the OP finances, engineers, and executes his/her own billion dollar spacecraft, they may place any instrument they like onboard ...and if that is your objective, start by reading my post on engineering tradeoffs in spaceship design, with links to a complete lecture-series... : but until then, accept at face-value this simple fact: somebody very smart thought about your idea and dismissed it for very good reasons, long before you even knew they were planning a Mars rover.Nimur (talk) 20:25, 9 February 2013 (UTC)[reply]
OMG, did you read anything that other people wrote before harping, falsely, about how it's impossible to record sound on Mars? NASA's Mars Polar Lander had a microphone, by design, for the purpose of recording sound. Phoenix carried a microphone by "accident", and though it heard nothing, the team thought that was because it got frozen and gave it a second try. The Planetary Society has built a Mars microphone for inclusion on a spacecraft, and it flew on the Polar Lander. Your claim that it's physically impossible to hear sound on Mars is simply wrong. --140.180.247.198 (talk) 00:40, 10 February 2013 (UTC)[reply]
I agree. Unusually for Nimur, he was wrong in detail (eg most modern cheap microphones are electret, not piezo - piezo has been obsolete for 30 years) and wrong in overall thrust. While the low altitude atmosphere density on Mars is only about 1/50th of that on on Earth, this is still quite sufficient to carry sound - you'll just need a much bigger microphone diaphram area to get equivalent sensitivity, as was explained in another recent question on Ref Desk. Detectable frequencies can easily cover a good bit of the audio range. However I also agree that nothing much is lost by not having the microphone. What do you expect to hear? I should think nothing other than the mechanical noises of the spacecraft, and a bit of wind noise. We know what wind noise sounds like. It's not as though we seriously expect the sounds of alien construction work or something. A transducer to measure wind velocity would be MUCH more cost effective. Wickwack 121.215.84.193 (talk) 03:08, 10 February 2013 (UTC)[reply]
I harp about many things, and so I am occasionally wrong. In this instance, I regret that my post did not cover all possible types of acoustic transducers; and that I did not formally reference or calculate the exact data that we would record for a particular sound pressure level at Mars' surface. My hope is that the interested reader could follow my line of reasoning and trivially estimate these parameters themselves, using their favorite type of transducer, amplifier, sampler, compression algorithm, and so forth. Perhaps this was an instance where the exercise should not have been left to the reader. As a result of this omission, I did not personally calculate the expected signal level, and so I may have overemphasized my position. In review, I concur that it is not "physically impossible" to record acoustic signals on Mars; but I stand by my position: recording acoustic signals on Mars is "still not an effective use of engineering time or science budget." I apologize that my overemphatic post may have inadvertantly conveyed an incorrect conclusion.
But, regarding electrets and piezos: it is a fair point that most new computers use condenser microphones; but piezos are not obsolete; my electric guitar was built in 2004 and uses piezo devices similar to this, and they sound quite nice. I have strong reason to believe that they are linear to many hundred octaves above the human ear's capability to hear sound (and far beyond the ordinary frequency range of a condenser microphone). And, in other work, I've used piezo couples for lots of reasons, particularly when I need to put my microphone in a vacuum, or on a rocket, or in a radiation-hazard enviroment, or any other place I need to doubt the integrity of a condenser microphone. Nimur (talk) 06:21, 10 February 2013 (UTC)[reply]
We are off-topic now, but you'll find if you check, those mikes in laptop computers are electret, not condenser. However the operating principle in both is the same (charge pumping by varying the distance between electrodes), and some people are sloppy with their terminology. In any case neither electret nor condensor are piezo. For electric guitars, magnetic pickups are used almost exclusively. Only in acoustic guitars with non-steel strings are piezo pickups (such as the Palathetic )used, as obviously magnetic pickups will not work without steel strings. For studio work, a conventional mike will give a better sound for acoustic guitars, though of course just what the sound is, is a matter of choice between the musician and the producer. Your link returns a 404 error. The design of condenser mikes is a compromise (a rather good compromise compared to other types of mike, but still a compromise), explained extremely well in Harry F Olson's, Acoustical Engineering - a venerable but still excellent standard text on the subject. Condensor mikes are certainly NOT good for hundreds of octaves above human capability - that would be greater than 15 kHz x 2100 ie ~2 x 1033 Hz, well and truely beyond the range of theoretical physics! Visible light is only around 6 x 1014 Hz! You are being ridiculous. Condenser mikes, by the nature of how they operate, are not inherently linear but typically are good covering just the frequency and amplitude range of human hearing. Wickwack 120.145.68.197 (talk) 09:37, 10 February 2013 (UTC)[reply]
It would violate the law of physics, i.e. "in space nothing can hear you scream". Gzuckier (talk) 05:13, 11 February 2013 (UTC) [reply]
Why is my charger emitting high pitched sound?
I think it has something to do with the transformer, but I don't know why. — Preceding unsigned comment added by Inspector (talk • contribs) 07:04, 9 February 2013 (UTC)[reply]
And I am guessing that the person who asked the question is either under 40 years old (if male) or under 60 years old (if female). Otherwise he/she would not be able to hear it. --Guy Macon (talk) 11:19, 9 February 2013 (UTC)[reply]
I guess the sound frequency is much lower. There is still one other strange thing when I studied further: when I connect it to my cell phone, the continuous buzzing stopped and changed to intermittent buzzing like heartbeat buzzing in the hospital.--Inspector (talk) 11:45, 9 February 2013 (UTC)[reply]
As far as I know this is an indication that it's either broken or or does not conform to the usual specifications required in industrialised countries (conformance is often claimed fraudulently). And that it's an incendiary risk. If it's expensive to replace, the device you are charging is not very valuable, and there is no reason to trust that a new charger will be better, then I personally would just continue to use it but would be careful never to keep it plugged in when I leave the house. Also, I would always think of it first when there are any problems with radio or tv reception, wlan etc. Hans Adler 11:52, 9 February 2013 (UTC)[reply]
Hans, I wouldn't go that far... it sounds like you're spreading a little FUD about a common, harmless annoyance. It's far more likely that the power-supply is functioning correctly, but its designers did not go through the painstaking analysis to guarantee that it never emits audible noise under any condition. If the device has a marking from UL - Underwriters Laboratory - or from the FCC, then its safety and emissions have been tested and verified, even if it makes an annoying noise. Outside the U.S., other agencies and industry groups are responsible for similar testing. This should not be construed to mean that any UL-tested product is always safe under any condition - you need to apply common sense and be aware of defects and damage. But, I am not aware of any requirement that audible noise in a switching power supply must be suppressed, and I don't understand why you think audible noise has any correlation to fire-hazard. Audible noise might be caused by any number of conditions, and most are not hazardous. In many scenarios, noise might indicate that the power supply is not operating at peak efficiency or that it's exerting unnecessary mechanical stress on some component, reducing its usable lifetime. But ... risk of fire? Surely if audible noise indicated fire-hazard, then every speaker cabinet would be a fire hazard during normal use? Nimur (talk) 19:46, 9 February 2013 (UTC)[reply]
A high-pitched buzzing noise might indicate sparking, but this is improbable. More likely this is the result of some transistor-induced harmonics causing the transformer core to intermittently vibrate and emit audible noise. 24.23.196.85 (talk) 20:27, 9 February 2013 (UTC)[reply]
I think Hans Adler is suggested that any device which emits noise must be a poorly made device which fails to comply with proper safety standards as present in most developed countries and any such markings suggesting otherwise are fraudulent so that the device must be a fire risk; not that noise itself indicates a fire risk. I agree this is going way too far. It may be slightly more likely a improperly designed or made power supply which violates safety standards makes audible noise, but there are a large number of perfectly safe devices which comply with whatever local standards which also make noise. And there will be plenty of unsafe devices which don't have noise.Nil Einne (talk) 21:18, 9 February 2013 (UTC)[reply]
Modern chargers are almost certainly switched-mode power supplies (SMPS). Our own article suggests such supplies emit acoustic noise (almost a tautology?) suggesting that such noise is "[u]sually inaudible to most humans, unless they ... are malfunctioning, ...". I think that means unless the SMPS is malfunctioning, not the human. Also "[t]he operating frequency of an unloaded SMPS is sometimes in the audible human range, and may sound subjectively quite loud for people who have hyperacusis in the relevant frequency range". QED? --Senra (talk) 21:55, 9 February 2013 (UTC)[reply]
I don't think I am alarmist, just a bit careful after I have read several reports like this one. Hans Adler 22:20, 9 February 2013 (UTC)[reply]
I have a bit of experience in this area, having designed everything from components for NASA manned space flight to toys for Mattel. Warnings are good, but they have to be accurate. Otherwise you are chicken little or the boy who cried wolf. The fact of the matter is that there is no known correlation between SPS acoustic noise and safety. You are warning people about the wrong things. --Guy Macon (talk) 23:26, 9 February 2013 (UTC)[reply]
I would also note that the example you gave (the counterfeit power supply with no components that fed the 12V straight through instead of the proper 5v) is completely silent and very unlikely to malfunction any more than it is already doing... --Guy Macon (talk) 00:56, 11 February 2013 (UTC)[reply]
It could be a phasor on overload, so toss it down the nearest disposal chute right away ! :-) StuRat (talk) 23:29, 9 February 2013 (UTC) [reply]
What! The resulting explosion might kill the princess! --Senra (talk) 17:22, 10 February 2013 (UTC) [reply]
Back to the topic, is it functioning right for the charger to change to periodical beeping when connected?--Inspector (talk) 02:02, 10 February 2013 (UTC)[reply]
It depends what you mean by beeping. If it is charging something that has charge control circuitry, the load might be switching on and off, but I wouldn't expect this to be a fast "beep". A slow cycling with the high frequency periodically going on and off might just be caused by the noise disappearing when current is drawn. This could well be normal operation because it is not unusual for the high frequency switching to induce a tiny vibration in laminations of the core, especially under no-current conditions. Technically, this would be regarded as a "fault" because energy is being wasted, but it is a very minor fault. If the charger is getting very hot, then there might be a more serious fault, but just warm is normal. Dbfirs 12:56, 10 February 2013 (UTC)[reply]
You have not told us the age and model of your cell-phone, nor whether the charger is the original or a low-cost replacement. If I had this problem, I would be searching the manufacturers forum to determine if others had reported similar problems. In your case, I would also eliminate other potential explanations. For example, responders above mention age and I noted hyperacusis as a possible cause. Have you asked non family members (in case hyperacusis is hereditary) if they can hear it too? --Senra (talk) 17:22, 10 February 2013 (UTC)[reply]
Are you sure this is not what you are actually hearing? μηδείς (talk) 21:33, 10 February 2013 (UTC)[reply]
i think i've heard that Leptin inhibits Gerlin. is that true?
If you only think you've heard something, it is unlikely to be true. Anyway, what is gerlin? Is it like trollin?--Shantavira|feed me 16:35, 9 February 2013 (UTC)[reply]
They are somewhat antagonistic, but it's not simple - our articles (note spelling) cover this pretty well: leptin and ghrelin. -- Scray (talk) 16:55, 9 February 2013 (UTC)[reply]
In a quick scan of our articles I didn't see any discussion of that question. There is some evidence that leptin inhibits ghrelin, for example PMID 15867335, but the data seem to be limited. Looie496 (talk) 16:59, 9 February 2013 (UTC)[reply]
...which is why I said what I said. -- Scray (talk) 02:55, 10 February 2013 (UTC)[reply]
in simple words
how do i culculate my bmi index?, please give me a verbal explanation. thanks 79.179.134.180 (talk) 16:32, 9 February 2013 (UTC)[reply]
Measure your height (in meters) and your weight (in kilograms), and plug them into this formula:
.
That's all there is to it. Looie496 (talk) 16:48, 9 February 2013 (UTC)[reply]
Weigh yourself, in kilograms. Measure your height, in metres. Square your height (that is, multiply your height by itself). So, if you are 1.5m tall, you do 1.5 times 1.5 (1.5 x 1.5). Make a note of your answer. Take your weight, and divide it by your previous answer. This number is your BMI. 86.163.209.18 (talk) 12:34, 10 February 2013 (UTC)[reply]
I think I have found a new way of synthesising graphite. In order to be certain that it is graphite, I need to know whether the route of production is a possibility.
Crystallisation of graphite from solution. I added organic material to 38% sulfuric acid, and boiled until the material was fully carbonised. The resulting product was a highly acidic, opaque, black sol (colloid)(?). Upon, cooling and standing for several hours, large, insoluble, opaque, black crystals have formed. I could send photos of the mystery crystals. Plasmic Physics (talk) 22:28, 9 February 2013 (UTC)[reply]
I suggest you try writing on paper with it, then using an eraser on it. If it acts just like graphite in both respects, it probably is graphite (lead behaves similarly, but I can't see how you could have produced lead without having put lead in). Of course, graphite is quite cheap, so you probably aren't going to make it any cheaper than the current methods. StuRat (talk) 22:31, 9 February 2013 (UTC)[reply]
It quite possibly is graphite. Adding oleum to sugar will produce a ball of carbon, that is larger that the reactants you started with. CS Miller (talk) 22:58, 9 February 2013 (UTC)[reply]
(edit conflict) The black stuff you made is the product of the dehydration reaction between the sulfuric acid and the organic material (a typical reaction for sulfuric acid, I should add), and as such, is essentially pure carbon. However, I should caution you that it might still contain free acid, which would tend to damage the paper. 24.23.196.85 (talk) 23:01, 9 February 2013 (UTC)[reply]
Considering that the sulfuric acid predominantly acts as a catalyst, and the reactant is essentially free, I fail to see how it would not be cheaper? Plasmic Physics (talk) 23:14, 9 February 2013 (UTC)[reply]
Well, the energy needed to boil it until carbonized isn't free. StuRat (talk) 23:31, 9 February 2013 (UTC)[reply]
Put it in perspective with the cost of mining, or generating thousands of degree temperatures for synthesis. Plasmic Physics (talk) 00:00, 10 February 2013 (UTC)[reply]
Should we ask the name of your unfortunate victim which you describe merely as "organic material" ? :-) StuRat (talk) 23:34, 9 February 2013 (UTC) [reply]
Or perhaps "Cooking Earl" ? StuRat (talk) 03:56, 10 February 2013 (UTC) [reply]
Is your organic material free? (don't forget transportation costs!) Is the energy used to boil the acid free? Is the amortized cost of the equipment free? Is the hazardous waste disposal that your washing step requires free? Labor? Renting a building? Remember, you are competing with people who are digging graphite out of the ground in China. --Guy Macon (talk) 23:39, 9 February 2013 (UTC)[reply]
According to the charcoal article, creating charcoal by the dehydration of sugar by sulfuric acid will produce the purest charcoal. CS Miller (talk) 23:45, 9 February 2013 (UTC)[reply]
Pond scum should work, 'organic material' is a pretty wide menu. As for waste desposal, such a plant could be adjacent to a sulfuric acid plant/phosphate fertilizer pant. The hardous waste could be recyled thus. Plasmic Physics (talk) 00:00, 10 February 2013 (UTC)[reply]
I've known people who work in the industry, so I feel safe saying that sulfuric acid production is nasty business. It produces gobs of really nasty waste which is a bitch to take care of, and we're not likely to be able to produce it in volumes necessary to produce enough fuel given that it is already, by far, the most produced chemical in the world. In order to make a viable source of carbon as a fuel to replace mined coal, you'd need more sulfuric acid than is availible. In 2011, the world produced 7,695.4 million tonnes of coal. That's 7,695,400,000,000 kilograms of coal. To produce a similar amount of carbon from random carbon sources, you'd need at least a 1:1 stoichiometric amount of sulfuric acid; probably a lot more. That'd be 7,695,400,000,000 *98/12 = 62,846,000,000,000 kilograms of sulfuric acid. The world produces about 180,000,000,000 kilograms per year, so even if you converted the entire world production of sulfuric acid over to your plan, you'd create enough pure carbon to replace 0.2% of the world's coal output. So, you're "cost of mining" issue is irrelevent. You couldn't possibly make enough carbon to make a difference in mining operations. --Jayron32 04:57, 10 February 2013 (UTC)[reply]
Just a note: this subdiscussion is about synthetic graphite, not coal. Plasmic Physics (talk) 05:47, 10 February 2013 (UTC)[reply]
You brought up the coal mining, not me. To what end are you proposing to use the graphite? Is there a shortage of graphite or is the economics of the current graphite market in sore need of you scaling up a middle-school classroom demonstration into an industrial process? --Jayron32 06:03, 10 February 2013 (UTC)[reply]
That was the first time I "coal" escaped my mouth, in a manner of speaking. Well, I think so. Plasmic Physics (talk) 08:46, 10 February 2013 (UTC)[reply]
Well, even if you were wanting to replace coal, it's not entirely out of the question. Firstly, you don't need to replace all coal, coking coal is the most valuable coal commodity, so extra costs of production are more easily absorbed there. Also, if you use a renewable source of organic material, then your material is carbon neutral (ignoring the energy input required which may or may nor be fossil fuel based), which gives it an advantage in a market with carbon trading or carbon taxes. Also, Jayron, who is clearly a very good physicist, is no economist. If there was a massive extra demand for sulfuric acid, there would be a massive incentive to increase supply. Unless we've already allocated every applicable resource to producing sulfuric acid, there's always more production possible. 202.155.85.18 (talk) 01:41, 11 February 2013 (UTC)[reply]
Unless we're already at or near capacity for producing sulfuric acid, then all we're likely to do is drive up the price or divert supplies away from other uses. --Jayron32 03:00, 11 February 2013 (UTC)[reply]
The price will be driven up by the increased demand. That is a given. You can only divert from other uses by offering a higher price. If we're at capacity for H2SO4 production, then we can only divert from other uses. 202.155.85.18 (talk) 03:42, 11 February 2013 (UTC)[reply]
(edit conflict) And what of the capital investment required for new sulfuric acid plants, which would run into the hundreds of megabucks? (Not to mention the other barriers to entry, like environmental regulations.) Also, graphite CANNOT replace coking coal -- coke must be not only chemically pure, but hard as well (because it must not crumble to dust when dropped a hundred feet or more down a blast furnace along with a tonload of iron ore and limestone), whereas graphite (no matter whether it's natural graphite, Acheson graphite or Plasmic graphite) is soft and easily abraded. 24.23.196.85 (talk) 03:06, 11 February 2013 (UTC)[reply]
Plasmic's exact phrase was "cost of mining" -- so he said "mining", not "coal". Graphite is a different material than coal, with different applications, but it's found naturally and mined in a similar manner to coal. It can also be produced synthetically by carbonization of coal tar pitch, which is more economical than Plasmic's method of dehydrating sugar or cooking oil with sulfuric acid (coal tar pitch being a major industrial waste from the steelmaking industry, and the raw materials for the Plasmic Process being more expensive, along with the prohibitive cost of hazardous waste disposal in the latter process). 24.23.196.85 (talk) 01:49, 11 February 2013 (UTC)[reply]
Raw material, as I've said, includes most organic material, even waste. Using cooking oil, or sugar would not be wise choices in this regard. Oil was merely used as a proof of concept. And sulfuric acid is not a main reactant, but a catalyst, it is ordinairily only slowly consumed. Plasmic Physics (talk) 02:47, 11 February 2013 (UTC)[reply]
It is gradually diluted and contaminated, and eventually must be disposed of as hazardous waste, every drop of it. Also, when the graphite is washed to remove the free acid (which would otherwise corrode anything that comes into contact with Plasmic graphite), the wastewater also has to be disposed of as hazardous waste. Sorry, the Acheson process is still much more economical than what you propose, even with the high energy costs these days. 24.23.196.85 (talk) 02:53, 11 February 2013 (UTC)[reply]
What are all of these supposed toxic byproducts? The sulpfuric acid itself ig gradually diluted with water, so you have dilute sulphuric acid which is great for making...concentrated sulfuric acid. Even if you did want to dispose of it, just add lime to balance the pH and discharge. 202.155.85.18 (talk) 03:39, 11 February 2013 (UTC)[reply]
This "dilute sulfuric acid" is actually UNFIT for making concentrated sulfuric acid, due to extensive contamination with organics and metals and whatnot. Purifying it before reconcentrating would be even MORE expensive than its disposal. 24.23.196.85 (talk) 04:15, 11 February 2013 (UTC)[reply]
It's not going to have "extensive contamination with organics". How do you remove organics from things? You percolate it through finely divided carbon. What was just in the acid? Oh yeah, finely divided carbon. And disposing of it is not expensive anyway. And there's no real need for the conc H2SO4 to be ultra pure anyway. The whole "recycling sulfuric acid" is not up for debate since it's a stable of industry, as in sulfuric acid recovery plants. 202.155.85.18 (talk) 04:35, 11 February 2013 (UTC)[reply]
Evaporate the liquid, purify and recycle the sulfurous oxides back into sulfuric acid. Using various separation techniques, the only waste should be in the form of a solid residue (metal oxides and sulfates). Plasmic Physics (talk) 03:41, 11 February 2013 (UTC)[reply]
And the energy needed for the purification and evaporation would cost how much? 24.23.196.85 (talk) 04:16, 11 February 2013 (UTC)[reply]
Without reviewing the chemistry, I think this sounds like a plausible notion to get coal out of biodiesel, but don't people actually want liquid fuels out of coal? Wnt (talk) 16:08, 11 February 2013 (UTC)[reply]
Would an electron and positron annihilating be considered two particles occupying the same space?GeeBIGS (talk) 02:21, 10 February 2013 (UTC)[reply]
Not after they were annihilated (which means they would have ceased to exist). -- Jack of Oz[Talk] 03:44, 10 February 2013 (UTC)[reply]
Well how about before or during then? Sounds like the same exact place to me....NOWHERE.GeeBIGS (talk) 06:20, 11 February 2013 (UTC)[reply]
No, both electrons and positrons are fermions which, by the Pauli exclusion principle cannot occupy the same quantum state (the concept of "same space" at this scale is somewhat meaningless; we don't speak in terms of location, since these particles are not localizable, but instead we speak of particles occupying a certain "quantum state", which is functionally equivalent to "location" for the purpose of discussions like this). However, bosons can occupy the same quantum state as each other; this leads to materials like Bose–Einstein condensates. --Jayron32 04:40, 10 February 2013 (UTC)[reply]
To my understanding, the Pauli exclusion principle only applies to the full quantum state of fermions. Any difference in quantum numbers will allow cohabitation - which is why you can get two electrons in one atomic orbital. They share quantum states except for one is "spin up" and the other is "spin down". I'm not familiar enough with QED to know the answer, but wouldn't the difference in charge on electrons and positrons be sufficient to introduce them into different quantum states, even if everything else about them was identical? If so, then the Pauli exclusion principle would say nothing about them regarding being able to occupy the same space or not. -- 205.175.124.30 (talk) 01:53, 11 February 2013 (UTC)[reply]
Well, that's why I hedged over the use of the term "same space". What happens is their probability space overlaps as they get closer, and eventually the likelihood of annihilation becomes great enough to cause it to happen, and it does. They can do this without the "points" occupying the "same space", or even having the same quantum state. --Jayron32 04:09, 11 February 2013 (UTC)[reply]
So how close do they get to each other before becoming empty space?68.36.148.100 (talk) 00:21, 11 February 2013 (UTC)[reply]
Electrons (and positrons) have no physical size, so are always empty space. (However they do have electrical fields which extend for a distance.) There is no specific cutoff distance for when they annihilate - the probability just gets larger and larger the closer they are. They can even orbit each over for a while, see: Positronium. Ariel. (talk) 00:59, 11 February 2013 (UTC)[reply]
Joke: how many electrons and positrons can you fit in a lightbulb? Infinity. But seriously, Could you give a little more in your explanation? Thanks.68.36.148.100 (talk) 01:32, 11 February 2013 (UTC)[reply]
What do you want to know? Ariel. (talk) 03:01, 11 February 2013 (UTC)[reply]
"their probability space overlaps as they get closer, and eventually the likelihood of annihilation becomes great enough to cause it to happen, and it does." I don't mean to sound rude , but this sound like some south Philly used car salesman triple talk to me. Ariel, could you please explain this to me in other words?GeeBIGS (talk) 06:15, 11 February 2013 (UTC)[reply]
An electron does not have a defined position, but rather there is a chance of finding it anywhere. Technically it could even be light years away from where you would normally think of as it's "location" - but at an extremely low probability. That's what he means by "probability space". So when the electron and positron approach each other there is a chance that they actually happen to both be in the same space at the same time, but also a chance that they are not "near" each other. The closer they get the greater the chance of finding them both in the same spot. Once that happens they will annihilate. The visible effect of this is time. Instead of annihilating instantly they take some time to do it, and you can define a "half life" of sorts of how long it takes them to both randomly happen to be in the same place at the same time. But remember that it's random - for the same distance some will annihilate extremely fast, some will take longer. But you can calculate how long it will take half of them to annihilate. And of course the closer they are the shorter that time is - but it's never 0 and it's never infinity. I hope this helps. Ariel. (talk) 09:10, 11 February 2013 (UTC)[reply]
Define "in the same spot" and "the same place at the same time" and if you could please explain how both simultaneously ceasing to exist is not the same as being in the same space . GeeBIGS (talk) 12:00, 11 February 2013 (UTC)[reply]
"the same place at the same time" means their waveforms overlap such that the waves exactly cancel out. You can (if you want) define "the same space" as the location of maximum probability. But it's not necessary for those two locations (as defined that way) to be the same for them to annihilate. Ariel. (talk) 19:01, 11 February 2013 (UTC)[reply]
No, they don't cancel, since that would violate conservation of energy. There's no special rule about annihilation in quantum field theory. Decays of the form particle + its antiparticle → other particle + its antiparticle are allowed (if energy-momentum is conserved) simply because antiparticles have opposite conserved charges, so all the charges add up to 0 on both sides and hence are conserved in the decay. In particular, the photon is its own antiparticle, so particle + antiparticle → 2 photons is always possible (though it may not happen very often if the particle interacts with light only indirectly). -- BenRG (talk) 19:41, 11 February 2013 (UTC)[reply]
Mass of a test tube full of bare nuclei
I read this in a book A test tube full of bare nuclei will weigh heavier than the earth. If the test tube is heavier than earth, it means test tube contains more nuclei than the earth. So, tube is heavier than the earth. This also means that the density of test tube is more than the density of neutron star. This doesn't sound good to me, I think the fact (italic sentence) is wrong. What do you think about this?
Show your knowledge (talk) 03:48, 10 February 2013 (UTC)[reply]
Are you sure you have that quote right ? It's bad English. Proper English would be "A test tube full of bare nuclei will weigh more than the Earth" or "A test tube full of bare nuclei will be heavier than the Earth". StuRat (talk) 03:54, 10 February 2013 (UTC)[reply]
I have written the same quote written in the book. Scientifically, I also think this quote is not right. What is your opinion about this? Show your knowledge (talk) 04:12, 10 February 2013 (UTC)[reply]
A test tube full of bare nuclei would be functionally as dense as a Neutron star, which has a density (at the lower estimate) of 3.7×1017 kg/m3 which means that a test tube full (say 5 cubic centimeters) would have a mass of 3.7×1017 kg/m3 x 5×10-6 m3 = 1.85 ×1012 kg. The earth has a mass of 5.9736×1024 kg, so no, it would not weigh more than the earth. It's 12 orders of magnitude lighter than the earth. But it's still really freaking heavy; a Boeing 787 has a mass of about 2.27×105 kg; so our test tube full of nuclei would still be heavier than 9,000,000 jumbo jets. --Jayron32 04:34, 10 February 2013 (UTC)[reply]
A typical test tube is actually more like 10cc's. But our Orders_of_magnitude_(mass) article (which is great for this kind of question) says that a teaspoonful (5ml) of neutron star material would have a mass of 5.5x1012...so I guess we might say 1013kg...but still nowhere near the mass of the earth. The orders of magnitude article suggest that a 1km tall mountain would have about the same mass as our test tube. About a thousand olympic-sized swimming pools full of nucleii would balance the mass of the Earth pretty well. SteveBaker (talk) 04:49, 10 February 2013 (UTC)[reply]
5.5... 1.85.... close enough for government work... ;) --Jayron32 05:06, 10 February 2013 (UTC)[reply]
So that quote is both wrong and bad English. I think I'd avoid that author. StuRat (talk) 04:50, 10 February 2013 (UTC)[reply]
I wonder if a pyrex test tube would be strong enough to contain a mass equivalent to millions of jet airplanes. ←Baseball BugsWhat's up, Doc?carrots→ 05:32, 10 February 2013 (UTC)[reply]
I wondered the same thing. I'm pretty sure the answer is no. That means we're talking about a strictly theoretical (= non-existent) test tube; and I don't see how we can have real nuclei sitting inside a non-existent test tube. So that must mean we're talking about theoretical (= non-existent) nuclei. In theory there's no difference between theory and practice; but in practice there kinda is, and this is a really good demonstration of it. -- Jack of Oz[Talk] 05:45, 10 February 2013 (UTC)[reply]
Well, it's not just the mass. What's keeping them squoze together? If they're "bare nuclei" other than neutrons, then the whole thing is positively charged and is furiously repelling itself while attracting surrounding electrons; I don't think you'd survive being anywhere near. If they're neutrons, then they have a half-life of about 15 minutes, meaning roughly 10000 times as radioactive as the Polonium-210 a microgram or so of which killed Litvinenko (in terms of disintegrations per unit time; admittedly each disintegration might be lower energy, but when you've got 10^26 times more of them I doubt that's going to help much). --Trovatore (talk) 08:21, 10 February 2013 (UTC)[reply]
Maybe only if it's really high quality pyrex. Although this strikes me as a half-cousin to the old question, "If you could develop a universal solvent, what would you store it in?" ←Baseball BugsWhat's up, Doc?carrots→ 06:21, 10 February 2013 (UTC)[reply]
This is great, I got the answer to my question in 1 hour. But I have already posted a question two days before (8 Feb, heading: Symbol of mass number) still I don't have the clear answer of that question. Please, do something about that. Show your knowledge (talk) 05:13, 10 February 2013 (UTC)[reply]
Sure, if you can get Wikipedia to double my pay for answering your questions, maybe I'll get on that. --Jayron32 05:34, 10 February 2013 (UTC)[reply]
Jayron32's pay rate is hereby doubled, so he will henceforth be paid twice as much per answer to Wikipedia Q as he was previously. StuRat (talk) 05:47, 10 February 2013 (UTC) [reply]
Strangely enough, my wallet doesn't seem any heavier. --Jayron32 06:01, 10 February 2013 (UTC)[reply]
In that case, I'd better triple your pay rate. StuRat (talk) 06:04, 10 February 2013 (UTC) [reply]
You'll get what's coming to you, I can guarantee you that, Jayron. -- Jack of Oz[Talk] 10:48, 10 February 2013 (UTC) [reply]
That's always what I am afraid of... --Jayron32 18:26, 10 February 2013 (UTC)[reply]
Isn't the issue here what exact meaning the word "full" has in this context? It talks about multiple nuclei, so that implies a least some spacing between them (if they were touching then wouldn't it just be one big nucleus?), and that spacing will by definiton determine the density. We normally say a test tube is full of water if any additional water would over flow, but we don't require the water to be compressed to qualify as being truly full. Conversely, a hydrogen cylinder is empty if it's pressure is equal to 1atm. 202.155.85.18 (talk) 00:31, 11 February 2013 (UTC)[reply]
I believe we're talking about the maximum density of neutrons it's possible to achieve outside of a black hole. Similarly to the core of a neutron star where the neutrons can't get any closer together due to quantum degeneracy pressure (ie the Pauli exclusion principle). True, our OP might be thinking of a less dense packing of neutrons - but what we're saying is that even at the densest packing imaginable, the test tube still only weighs as much as Snowdon - a modest-sized mountain. SteveBaker (talk) 16:04, 11 February 2013 (UTC)[reply]
I just realized...if you turned something with the mass of the Earth into a black hole, it would have a Schwartzchild Radius of about 1cm. That's a 2cm diameter - so at that density, it would just about fit into a large test tube...but not into a smaller one. SteveBaker (talk) 16:09, 11 February 2013 (UTC)[reply]
solar ac dc inverters failure
Is there a way to test an inverter to see if it is functioning properly, or do they just quit completely? My batteries are not lasting like they should and was told it could be the inverter. off grid207.212.113.253 (talk) 04:03, 10 February 2013 (UTC)[reply]
As a general, but not absolute, rule, electronics tend to fail catastrophically, or fail with some part getting far too hot, emitting a "brown smell". So, if your invertor has not failed completely, is not showing any warning/fault display, and is not emitting burnt smells, it's most likely not your problem. You can test an invertor by testing first with no load, then with something (such as an electric room heater) that loads it to at or near its full rating. Measure the DC input current for both conditions, and comnpare with the manufactuer's data. Typically the DC input amps should be roughly 1 twentieth the rated load in watts divided by the battery voltage. The full load DC input amps should be about 1.1 times the actual load divided by the battery voltage. Do not be concerned if the no load DC amps varies somewhat from this.
Are you aware that batteries are subject to both catastrophic failure and gradual loss in capacity? Battery life is dependent on brand quality, ambient temperature, the number of cycles, and the depth of discharge. In the hot climate regions of Australia, battery life for cheap European lead acid batteries can be as short as two summers. A life of ten years for a generously sized quality Janpanese battery is a good achievement. The output of the cheaper solar panels also deteriorates over time. A panel service life to 50% output of 10 years is not untypical. Sometimes the sealing fails and water gets into the silicon, causing an early drop in output. Keit 121.215.141.120 (talk) 05:50, 10 February 2013 (UTC)[reply]
In addition to Keit's excellent answer above, if you need to replace your batteries, you should be aware that the batteries sold for vehicle use are usually not suitable for long-term use with an inverter because they are not designed for deep discharge. Marine type batteries, or ones designed for your specific application should last much longer, but will cost at least twice as much (in the UK, at least). Dbfirs 08:40, 10 February 2013 (UTC)[reply]
IIRC, the "never add water" batteries which have become ubiquitous for automobiles these days (contain calcium, I believe) are much more susceptible to damage by deep discharge than the old "check and fill periodically" variety; my personal experience is that one instance of leaving the headlights on overnight or similar can make them essentially "dead", i.e. they may work under ideal circumstances but leave them for a couple of weeks, or the first cold day, and no juice. So like the guy says, don't use car batteries. Gzuckier (talk) 05:22, 11 February 2013 (UTC)[reply]
Do photons produce gravitational field ?
I know photons produce electric as well as magnetic field. I want to know, do they produce gravitational field ? I know electric field is always accompanied by magnetic field. Is it correct to say that "magnetic field is also always accompanied by electric field" ? There is another confusion, suppose if we increase the frequency of a photon, its energy also increases. If we increase the wavelength of a photon, its energy decreases. Since, E = hv. Give your response about last three sentences. Thanks! Parimal Kumar Singh (talk) 04:07, 10 February 2013 (UTC)[reply]
Photons certainly interact with gravitational fields, gravitational lensing, for example. Also, not every electric field necessarily produces a magnetic field. A magnetic field is generated by a moving electric charge. A stationary electric charge produces an electric field, but no magnetic field. I'm not sure I follow your confusion on your last three sentences. Frequency and wavelength are inversely proportional: if you get more vibrations, the space between them gets shorter; if you get less vibrations, the space between them is longer. So decreasing energy = decreasing frequency = increasing wavelength and vice-versa. --Jayron32 05:05, 10 February 2013 (UTC)[reply]
If an electric field is static, i.e., not moving or changing in strength, there is no resultant magnetic field as Jayron said. If the electric field is not static, there still will not be a magnetic field unless the field encloses at least part of an electric circuit - the magnetic field arises from the current in the circuit. Conversely, a magnetic field will not create an electric field if the space is a perfect electric conductor. In electromagnetic radiation, electric and magnetic fields propagate together. Wickwack 124.182.176.205 (talk) 05:36, 10 February 2013 (UTC)[reply]
Photons contribute to the energy density of a volume, which could be included in a generalized energy tensor used to calculate behavior using general relativity. In most scenarios, the effect is negligible because the energy-density of photons has a much smaller magnitude than the mass density of ordinary matter. Nimur (talk) 06:07, 10 February 2013 (UTC)[reply]
When I was reading about Albert Einstein, I found that photons have attraction towards gravitational field. Earth has mass and it produces gravitational field. Do photons produce gravitational field like our earth? Parimal Kumar Singh (talk) 10:54, 10 February 2013 (UTC)[reply]
Photons are massless, so they don't produce gravitational fields.Dja1979 (talk) 12:00, 10 February 2013 (UTC)[reply]
See Stress–energy tensor. Radiation can indeed be a source of gravitational fields in General Relativity. Jheald (talk) 12:20, 10 February 2013 (UTC)[reply]
They do produce a gravitational field due to the energy they have which depends on their frequency. If you had a box with mirrors it would weigh more if it had some light being reflected around inside it. Dmcq (talk) 12:25, 10 February 2013 (UTC)[reply]
To elaborate a bit on the above answers, note that one can always choose a reference frame such that an electron's magnetic field "disappears". It still exists however in other reference frames. Furthermore, the electric and magnetic fields are mediated by the energy of photons, the precise value of which is frame dependent in accordance with E = hv and the relativistic Doppler effect. Also important is the fact that mass and energy are equivalent and since gravity is a consequence of mass-energy, light produces gravity and it is also affected by it. -Modocc (talk) 14:41, 10 February 2013 (UTC)[reply]
This is a hugely perennial question at the Refdesk and just from the first 20 search results for "photon" and "gravity" I find [19][20][21] (also [22] which is useless). Of these, [23] appears to be the useful answer if you can access it. At some point it would be great if someone could write an actual article about photon gravity explaining this and any other good sources so we can send people there. There's some unexpected factor of 2 involved. Wnt (talk) 16:03, 11 February 2013 (UTC)[reply]
It's probably related to the factor-of-2 difference in the deflection of light by the sun, which is one of the most famous tests of general relativity. If so, it's more or less because spatial and temporal curvature contribute equally to the deflection when light is involved, while with nonrelativistic particles (and in the Newtonian limit) only the temporal curvature matters. I think. -- BenRG (talk) 19:25, 11 February 2013 (UTC)[reply]
Planets and frequency of sound
Dear Sir, this is not my homework
1) Which planet of solar system-
a) rotate faster
b) rotate slower
c) revolve around sun faster
d) revolve around sun slowerthan any other planet?
2) If i hear a sound of frequency 19,000 Hz, will it cause pain to my ears? What is relation between frequency and decibel?
You will find the answers to 1) in the article Planet#Planetary_attributes. Please let us know if you need help interpreting this. Frequency is the pitch of a note, and 19 kHz is at the top of the hearing range, in fact most of the population cannot hear a note this high. If you are young, then maybe you can, especially if it is loud. Decibels are a measure of sound intensity or loudness. See the article Decibel for technical details. Any very loud sound can be perceived as pain and may cause hearing damage. Dbfirs 08:27, 10 February 2013 (UTC)[reply]
Agreed, and let me add one more tidbit. It takes less energy to produce high frequency sounds of a given volume than low frequency sounds of the same volume. Therefore, say 20 watts at 19 kHz may very well be painful, while 20 watts at at a lower frequency is not. StuRat (talk) 03:57, 11 February 2013 (UTC)[reply]
Tensor
I was searching for a simple definition of tensor, I tried to read the linked article, but it was beyond my knowledge. Then, a professor told me a simple definition - A physical quantity is said to be tensor if it is neither a scalar nor a vector as its direction is not properly specified, but has different values in different directions. Examples of tensor include strain, moment of inertia, density, refractive index, etc. Wikipedia article on tensor says just opposite - Vectors and scalars themselves are also tensors. Who is correct? Which one should I prefer? Sunny Singh 07:57, 10 February 2013 (UTC) — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs)
Oh, you've stepped in it now :-)
The simple answer to your question is that scalars and vectors are tensors, of rank 0 and 1 respectively, but you don't normally call them that, because why bother? Your professor was explaining tensors of rank 2 and beyond. --Trovatore (talk) 08:00, 10 February 2013 (UTC)[reply]
You might enjoy Dan Fleisch's "What's a Tensor?" video in which he uses "children's blocks, small arrows, a couple of pieces of cardboard and a pointed stick".[24]Sean.hoyland - talk 13:54, 10 February 2013 (UTC)[reply]
I'm biased due to my computer science background, but I think the simplest way to think about a tensor is as a multidimensional array. Imagine a one-dimensional array, say arr[32]. In mathematics, this represents a vector, which is a rank-1 tensor. Imagine a two-dimensional array, arr[32][32]. This is a rank 2 tensor, and every value arr[i][j] presumably has some physical meaning specific to the indices i and j. Similarly, arr[32][32][32] would be a 3-dimensional array and hence a rank-3 tensor, etc.
Here's a concrete example to help with your intuition. The stress-energy tensor describes the movement of four-momentum through spacetime. It's a rank 2 tensor, so you can imagine it as the array T[4][4]; it's 4x4 because spacetime has 4 dimensions. T[i][j], the jth column of the ith row of the array, is the flux of the ith component of momentum through the jth dimension. For example, since by convention the 1st dimension is x, T[1][1] is the flux of the momentum in the x direction, across the surface defined by constant x. Similarly, T[1][2] is the flux of the momentum in the x direction, across the surface defined by constant y. --140.180.243.51 (talk) 19:38, 10 February 2013 (UTC)[reply]
The problem with that formulation is that it's coordinate-specific (or basis-specific, depending on whether you're thinking of physics-style or algebra-style tensors).
To me a rank-two tensor is a machine that takes one vector and gives you back another. For the stress tensor, you can think about the input vector being the one that describes a little surface of a place you could potentially put a cut inside the object, and the output vector is how much pressure or shear the object puts on that surface. None of this needs any coordinate system to be specified, and you don't need to break it down in to components to understand it. (When you want to make a practical calculation, you probably will specify a coordinate system and take components, but you'll use whichever one is most convenient at the time rather than one specified in advance.) --Trovatore (talk) 23:50, 10 February 2013 (UTC)[reply]
Hydrogen Synthesization
Can hydrogen be synthesized by manipulation of subatomic particles? — Preceding unsigned comment added by Lawrie1 (talk • contribs) 13:56, 10 February 2013 (UTC)[reply]
Sure. Take one proton and one electron, bring them into contact: presto, you have a hydrogen atom. Looie496 (talk) 16:17, 10 February 2013 (UTC)[reply]
All of the science of chemistry could be described as "manipulation of subatomic particles" - but as Looie says, we can also do it directly from a stream of protons and a stream of electrons. SteveBaker (talk) 15:53, 11 February 2013 (UTC)[reply]
Power lines: buried or overground?
What are the advantages of one system or the other? Is it just the cost of implementing vs. the cost of maintaining? OsmanRF34 (talk) 16:06, 10 February 2013 (UTC)[reply]
It depends on how you define "advantages". Ruslik_Zero 16:47, 10 February 2013 (UTC)[reply]
"Underground cables take up less right-of-way than overhead lines, have lower visibility, and are less affected by bad weather. However, costs of insulated cable and excavation are much higher than overhead construction. Faults in buried transmission lines take longer to locate and repair. Underground lines are strictly limited by their thermal capacity, which permits less overload or re-rating than overhead lines. Long underground cables have significant capacitance, which may reduce their ability to provide useful power to loads." See our article Electric power transmission, also Overhead power line and Undergrounding. Alansplodge (talk) 16:52, 10 February 2013 (UTC)[reply]
There are also safety considerations that generally favor underground lines. Looie496 (talk) 16:57, 10 February 2013 (UTC)[reply]
Also note that there are two variants of underground power lines, those with access tunnels and those without. Most seem to lack access tunnels, and even go under streets, etc., requiring digging up those streets to access them. Those with access tunnels, while more expensive initially, make for far easier, quicker, and cheaper maintenance, upgrades, etc.
And, in the comparison of above ground to underground, we can't neglect that most people find above ground wires to be ugly. As such, they may bring down property values. StuRat (talk) 04:03, 11 February 2013 (UTC)[reply]
True, but the higher construction and maintenance costs of underground cables still appear to trump everything else. 24.23.196.85 (talk) 04:11, 11 February 2013 (UTC)[reply]
That depends very much on local conditions. In Germany, overland lines are above ground (as mentioned before, if you distribute AC, underground has significantly larger losses over long distance), but local distribution is nearly exclusively underground. It must be 30 years or so since I last saw an overhead line go into a private building. The cost advantage shifts based on population density, and also on Quality of Service. --Stephan Schulz (talk) 12:48, 11 February 2013 (UTC)[reply]
Breakdown of a Barrel of Oil
Oil is used to produce fuels and petrochemicals. I believe that more of the oil is used for petrochemicals and for fuels, and that an oil company's profits are derived more from petrochemicals than from fuels. Is there an entry that provides the (approximate) breakdown? 68.54.32.39 (talk) 17:39, 10 February 2013 (UTC)[reply]
A breakdown of the products made from a typical barrel of US oil.
Crude oil is what's taken from the ground and comes in barrels; it's described as sweet if it has low sulfur content which makes it easier to work with. The article gives average contents. It is then refined by heating and distillation and various other chemical processes that separate out and create different petrochemicals. So it's not so much that they just separate out what's already there as it is what they make from it. μηδείς (talk) 18:03, 10 February 2013 (UTC)[reply]
Most refineries use distillation to separate the crude oil into different length hydrocarbons. They also use fluid catalytic cracking to split long-chain molecules into more-in-demand shorter ones. It is also possible to merge short-chained hydrocarbons into longer ones, which might be useful if there is more methane that there is a demand for. CS Miller (talk) 19:44, 10 February 2013 (UTC)[reply]
I've added the article caption to the chart to make clear that the product listed are not the unrefined makeup of a barrel. I'm also moving this up to my post to save space--if you object, go ahead and revert the edit. μηδείς (talk) 03:54, 11 February 2013 (UTC)[reply]
Thanks for that by the way. Good call. --Jayron32 04:05, 11 February 2013 (UTC)[reply]
What are those percentages though? Percent by volume? By weight? By energy content? Either way, that wouldn't answer the OP's question which is "By dollar value". The original WP:RS upon which that chart is based [25] is similarly unclear. Any chart with unlabelled axes is essentially useless!! SteveBaker (talk) 15:44, 11 February 2013 (UTC)[reply]
<-I'm not sure the premise of this question is entirely valid. It's probably fair to say that to a first approximation, oil company income comes from finding and producing hydrocarbons (i.e. their "upstream" actvities) rather than doing things to those hydrocarbons (i.e. their "downstream" activities). Obviously it's a lot more complicated than that in reality and each company is different in terms of the upstream vs downstream activities/profitability, but Exxon Mobil earnings for 2011 after tax for example were ~41 billion dollars, ~34 of which came from their upstream operations. Sean.hoyland - talk 04:27, 11 February 2013 (UTC)[reply]
Upright exercise bike question
What should be the position of the knees relative to the feet while riding an upright stationary bike? I mean should the knees be in the same level of the feet (distance between feet = distance between knees), or the knees be slightly bent outwards (distance between feet < distance between knees)? Which is the correct posture to prevent knee injury? Is 18 inches distance between the outer edges of the the two feet when they the over the pedals a safe distance? --PlanetEditor (talk) 17:47, 10 February 2013 (UTC)[reply]
The knees should be in the same plane as the feet, as nearly as possible. I can't speak to specific distances, but generally you want your knee bending in the way it naturally bends, not in any other direction. Riding significantly bowlegged is bad. Looie496 (talk) 19:44, 10 February 2013 (UTC)[reply]
After asteroid 2012 DA14 passes by the Earth in a few days, its orbital period will be 317 days. That is very close to a 7:6 ratio with Earth's orbital period. Is this significant - i.e. an orbital resonance? Bubba73You talkin' to me? 19:39, 10 February 2013 (UTC)[reply]
A resonance is a result of many small perturbations accumulating to give a stable relationship. In this case the 317 day period will result from a single huge perturbation, the previous period being 368 days, so it couldn't possibly be a resonance. Generally speaking Earth's gravity is too weak in comparison to Jupiter's for Earth to be able to produce stable orbital resonances. Looie496 (talk) 19:51, 10 February 2013 (UTC)[reply]
The exception being resonances with very low denominators - particularly 1:1 resonances. See horseshoe orbit for instance. --Tango (talk) 19:00, 11 February 2013 (UTC)[reply]
Dr. Imenson (spelling?) medical doctor
Dr. Imeson (not sure of the spelling) was my family doctor in the late 1940's and into the 1950's. I would like any information on him. What happened to him. When he passed on, etc.. — Preceding unsigned comment added by 24.7.138.9 (talk • contribs) 17:25, February 10, 2013
I need to add that he was a doctor in San Francisco, California — Preceding unsigned comment added by 24.7.138.9 (talk • contribs) 17:28, February 10, 2013
There are plenty of people-search websites you could try. A normal Google search turns up the following possibility:
Dr. Shale Imeson, MD Anesthesiologist, Pain Management Physician in Stockton, CA. However, since he's still around, it is unlikely that he was a Dr. in the '40s-'50s -- Maybe his father? ~:74.60.29.141 (talk) 23:46, 10 February 2013 (UTC)[reply]
February 11
Anodisation of Aluminium
Ive read numerous stories of iphone 5s being scuffed from the moment they opened the box. Ive also heard similar stories with other anodized aluminium products. This leads me to wonder, is anodisation a process which is difficult to achieve without causing small scuffs or marks? Clover345 (talk) 00:13, 11 February 2013 (UTC)[reply]
Not at all, but aluminum is a relatively soft metal, compared to say, stainless steel which I think was used on some of the earlier iPhone models. Metals of high hardness scratch less easily (and a scuff is basically just a scratch as I understand the term). 202.155.85.18 (talk) 00:17, 11 February 2013 (UTC)[reply]
One of the advantages of traditional anodising done with electric current per standards is that it provides a surface significantly harder than the base aluminium. However, it is a porus surface, further treatment (eg varnish) is usually done if the device is to be handled. See http://en.wikipedia.org/wiki/Anodizing. There could be a production problem with the further treatment. Also, there are various non-electric processes that result in what looks like an undyed anodised surface - these are not as tough. A substitute finish used by a company I once worked for used the following (much cheaper) process: Step 1 - clean with detergent and air blower dry. Step 2 - dip in dilute caustic soda. Step 3 - clean again with citrus solvent (slightly acidic). Step 4 - water rinse and air blower dry. Step 5 light spray with thinned marine varnish. The result looks identical with genuine anodising, but has no where near the toughness and durability - and usually the product left the shop with marks already on it. Ratbone 121.215.57.135 (talk) 01:06, 11 February 2013 (UTC)[reply]
Aluminum is relatively soft, but an anodized surface on aluminum is very, very hard. this page claims that apple uses something called "soft anodizing" which I am unfamilar with. --Guy Macon (talk)
Betretta M9 hammer hole
What is the purpose of the hole in the hammer spur of the M9 Baretta? 202.155.85.18 (talk) 02:42, 11 February 2013 (UTC)[reply]
Having less weight on the hammer means it can move faster, reducing the time between pulling the trigger having the gun fire. I think it is also done for aesthetic reasons. --T H F S W (T·C·E) 04:54, 11 February 2013 (UTC)[reply]
Maybe it's a design feature to match the seemingly useless hole at the bottom of the handle. ←Baseball BugsWhat's up, Doc?carrots→ 12:09, 11 February 2013 (UTC)[reply]
Not necessarily useless - ISTR some soldiers (or rather, officers) used to have a piece of string attaching their handgun to their belt (or holster, or similar). The hole at the bottom of the handle is in the perfect place for doing this. --Demiurge1000 (talk) 12:48, 11 February 2013 (UTC)[reply]
It's called a "skeletonized hammer". This means the gun has the same hammer geometry as it would with a solid hammer, but the hammer is lighter. Beretta themselves say here that this "can improve cycle time because of the lighter weight." Alternatively, as the hammer is lighter, you can replace the hammer spring with a weaker one (a lighter hammer is less mass for the spring to accelerate). This gives the gun a lower "pull weight" (the amount of force needed to pull the trigger); a smaller shooter (with weaker fingers) might opt for a lower pull weight, so they can still operate the gun easily. Naturally a lower pull weight isn't without issue, as a more sensitive trigger can make it more likely that the gun is fired accidentally. One can also get skeletonized triggers, which again lower the mass of the moving trigger system. -- Finlay McWalterჷTalk 16:23, 11 February 2013 (UTC)[reply]
hair drier off lithium batteries?
can a hairdryer (very high draw device) be run off of any common, relatively light-weight batteries? The usage case would be that it is OK to charge for a long time, but then a very high drain hairdrier should deplete it in just a minute or two... are any battery technologies appropriate? If the battery would get warm, the hairdrier can actually draw air from over it (preheating the air) so maybe this helps a bit. The application is actually on an RC helicopter that is supposed to work in a very new way. The battery for the hairdrier can be separate. Note: this is just about a special application, like an rc helicopter that can help glue sheet paneling. Thanks! 91.120.48.242 (talk) 12:32, 11 February 2013 (UTC)[reply]
Uh, this is a pretty resounding "no". Anyway, nobody uses a hairdrier for "just a few seconds" so I'm wondering if you can answer my quesiton more directly. (The hairdrier was a stand-in anyway.) Can a battery release 1600 watts for, say 30 seconds at a time, without damage or terrible overheating? I mean, long-charge, fast-discharge batteries... does that describe the profile of existing, compact, light-weight batteries? thanks... 91.120.48.242 (talk) 13:33, 11 February 2013 (UTC)[reply]
Probably not; a hair drier pulls an intense amount of current, and there's just not enough electrons at a high enough energy in a small, compact lithium ion battery to power such a devise for a meaningful amount of time. Battery-operated hair driers could possibly run off of the sorts of rechargable batteries used for cordless drills and tools like that, but only if you were just running a powerful fan motor. The heating element is the real energy hog in the hair drier, and you really need a huge amount of electrical power to run such a heating element effectively. The chemistry just doesn't work out given the amount of material available in any compact chemical battery. --Jayron32 16:41, 11 February 2013 (UTC)[reply]
RC hobbyists sometimes want fairly high discharge rates and they usually rely on lithium polymer batteries for the purpose although what you're asking for seems a bit extreme to me. It may be barely doable but it depends greatly on what you mean by 'relatively lightweight'. For example our article mentions some batteries managing up to 65C [26] continuous. As an example it links to [27]. If the specs are correct, this battery can handle up to 292.5A continuous discharge. With the nominal voltage of 37V (it's 10S) this means it should be able to supply 10,822.5W although given the rate of 65C this will be for less then a minute. However the battery itself is over 1.3kg so may not be considered lightweight. (I have to admit I wonder if the cabling and battery is really capable of handling 292.5A but perhaps it can.) Of course you don't need quite that level of power. So if we look under the category and sort by weight, you can find [28] which is 2200 mAh which given the rate of 65C can handle up to 143A continuous discharge. With the nominal voltage of 11.1V (it's 3S), this will be 1,587.3W so may fit your purposes. The battery is 266g. Of course you don't really need such a high discharge rate, more cells in series with a lower discharge rate would be fine. For example you could also use [29] which at 6S and 40C is supposed to be able to handle up to 72A continuous discharge so should be able to supply 1,598.4W for slightly longer then a minute although it's heavier at 385g. Probably one problem is generally for RC purposes, if you want a high voltage it's generally for a bigger device so a heavier battery isn't such a big deal. Of course the other point is even at 10S, you're still going to need about 43A. Personally I wouldn't want a lithium polymer battery supplying 43A near me (and if it's supplying 43A to a hair dryer it's likely to be near me) or I may find my hair getting dry a lot fast then I intended, along with the rest of my body. (While RC batteries are subject to shocks a hairdryer hopefully won't be, there are plenty of videos showing what can go wrong.) Nil Einne (talk) 17:00, 11 February 2013 (UTC)[reply]
It sounds like you don't actually want a hairdryer for heating hair but even so I'd suggest that even if it's theoretically possible, it's not practical for your purposes. These sort of things are at the bleeding edge (particularly when pushed close to the limits of the specs) and are targeted at a hard core group of hobbyists willing to take the associated risks (and as I said, there are plenty of videos and forums posts to show it). And in fact given your purposes it sounds like you will be in the group where 'rough handling' does come in to play. And since it sounds like this has some sort of industrial purpose, I quite doubt OSH and the fire brigade or the equivalent in your jurisdiction will be happy about the occasional exploding battery while charging, landing or storage (after a crash for example). Nil Einne (talk) 17:22, 11 February 2013 (UTC)[reply]
A look in RC forums suggest most 'C' ratings are, shall we say, optimistic, particularly when coming from HK and Chinese sellers. However this rather old post [30] suggests 40C may be possible or even 46C (120A in the case in point) although that was only with a single cell of a 3S battery. Of course even if all 3 could supply that at the same time you'd still be about 400W short, you'd need 4S. I didn't find the weight but it was suggested to be typical given the capacity (3S, 2600mAh) so from what I saw earlier it'll probably be under 400g.
In any case, I'd note you only seem to be considering this 'hair dryer' load. If you're planning some sort of RC helicopter, you'd also need power to lift the helicopter itself along with the battery and !hair dryer, take it to the destination, stay there while you glue the boards and then land again. It sounds quite unlikely 30s will be enough for this, so you'd need to adjust your demands and requirements. Presuming you come up with something like 3-5 minutes total flying time and you don't need more then say 300 watts for the helicopter (including all loads), from what I've already seen from over 1.5 years ago, it does sound likely it may be possible. I still think it isn't practical for anything other then demonstration/we can do this purposes. Edit: And I also question what you'll actually achieve if you only heat the glue for 30 seconds and then need to charge your helicopter or send a new one. Notice also the comments on cycle life and heat.
A discharge rate that fast might be closer to the capabilities of a capacitor than a battery, specifically a motor capacitor. Note that such devices are dangerous, as the voltages produced may match or exceed mains voltage. StuRat (talk) 16:33, 11 February 2013 (UTC)[reply]
Sturat, this is very good/interesting. I didn't know this before. Do you think it's possible to have one that is really very much "capped" (hehe) at 1600Watts, and will discharge for, say, a full 1 minute, if it has 1 minute * 1600 Watts in it for the voltage that's being drawn? Thanks. 91.120.48.242 (talk) 16:43, 11 February 2013 (UTC)[reply]
Stu is correct that a capacitor would provide a large current for a short time, but you are very unlikely to find a capacitor that you can afford with enough capacity to run a hair dryer heating element for more than a second or two. A very big capacitor might run a small DC fan motor for a minute or so, but would gradually reduce in speed as the capacitor discharged. I'd advise running a 12v hair dryer from a car booster battery, not lithium. Dbfirs 17:09, 11 February 2013 (UTC)[reply]
Nil's suggestion above, of a lithium polymer battery, sounds like it suits your purposes better than a capacitor, although it's still dangerous. StuRat (talk) 17:32, 11 February 2013 (UTC)[reply]
It would help if you could describe the purpose of this device more. Are you trying to dry paint on an high wall ? If so, perhaps shining lots of heat lamps at it from the ground might be a better approach. StuRat (talk) 17:38, 11 February 2013 (UTC)[reply]
I think you need to channel the antimatter emitter directly into the output stream... no, really, what I'd wonder is, is there a way that you can store the energy chemically in the glue itself, rather than electrically? But I like the direction of your mind. :) Wnt (talk) 17:40, 11 February 2013 (UTC)[reply]
From the OP's previous posts, I do wonder the state of the OPs mind when they had this idea. Nil Einne (talk) 18:21, 11 February 2013 (UTC)[reply]
Resonant frequency of an aluminium bar.
Hi! I'm trying to get a rough idea of the resonant frequency of an aluminium bar that's 1400 x 40 x 40mm and can probably be considered to be "unclamped" - in the plane that I'm expecting the vibration to occur.
I don't need an exact answer - I'm just trying to find out whether it's likely to have a resonant frequency somewhere between 10 and 20Hz.
Hmmm. Looks like the speed of sound in aluminum is roughly 5100 m/s. So sound travelling 1.4 m and back should take 2.8 m / 5100 m/s = 1800 Hz. The other directions should be higher. Acoustic resonance says more about this. My perception is that such disturbing "infra"sound can come from ducts less than twice this diameter, but this is presumably because of some lengthwise geometry, that the speed of sound in air is 15 times slower and perhaps because for some reason the formula for a closed tube is /4L instead of 2L. Wnt (talk) 15:50, 11 February 2013 (UTC)[reply]
Atom Volume
I can't seem to find any references for the assertion that "approx. 99.9% of an atom's volume is empty space". Can someone please provide some references, or is this assertion not verified? Thanks.165.212.189.187 (talk) 16:55, 11 February 2013 (UTC)[reply]
Which article says that? Dmcq (talk) 17:58, 11 February 2013 (UTC)[reply]
I googled 99.9% of an atom's volume is empty space and got a lot of hits, some of which at a glance seem to be legit. Duoduoduo (talk) 18:23, 11 February 2013 (UTC)[reply]
It depends on what is meant by "empty space". If you consider electrons to be point particles, like little tiny infinitesimal balls, and the nucleus to be another tiny, solid ball, then I suppose the claim could be made. But that's a rather primitive and not-very-accurate view of what an atom is. Instead, if you consider an atom to be a nucleus surrounded by the probability space which contains the electrons, then it's as solid as anything; if by solid you mean "impenetrable little hard nugget", then an atom is essentially solid, and none of it is "empty". The entire volume of any given atom is essentially all made up of the electron cloud, which is essentially impenetrable by other atoms. So the "claim" that atoms are "empty space" is based on some rather silly leaps of logic from a non-too-accurate view of what an atom is. It sounds all profound and all, but its basically bullshit. --Jayron32 18:41, 11 February 2013 (UTC)[reply]
You can take it the other way too. At atom can be considered 100% empty space. The electrons are point like particles with no size, the nucleons are in turn made up of quarks which are also point like. So in some way an atom is 100% empty space. Of course in the real world atoms interact not by their location, but by the forces acting on them and those forces have ranges. So the forces acting on the electrons certainly reach the nucleon, and the entire area in between can be considered to be "filled in" with that force. Same for the strong force holding the nucleons together and also for the force in between the quarks. The interesting conclusion is that the "size" of particle depends on what force you want to use to measure it with. Ariel. (talk) 18:51, 11 February 2013 (UTC)[reply]
That's why any visualization you can conceive of to represent the atom or its parts always falls woefully short of how it actually behaves. See the discussion a few days ago regarding this very topic. --Jayron32 19:29, 11 February 2013 (UTC)[reply]
Is this a true statement?: Things that don't exist have the same quantum state.165.212.189.187 (talk) 17:00, 11 February 2013 (UTC)[reply]
If coded in Predicate logic and following its usual axioms then that would evaluate as true. Basically what we'd be asking is whether the statement for all x and y, do x does not exist and y does not exist imply property 'quantum state the same', is that falsifiable ie can we find x and y so it is provably false? However this is an empty meaningless statement and the answer I gave is one more suited to the maths reference desk. Dmcq (talk) 17:49, 11 February 2013 (UTC)[reply]
More concisely: Everything exists, so any sentence starting with things that don't exist have... is vacuously true. Falisfiability and provability are unnecessary distractions. --Trovatore (talk) 18:59, 11 February 2013 (UTC)[reply]
Yes, that's the heart of it. We can also have fun with the vacuous truth of conditionals that have false antecedants. For example this sentence: "If this question is posted in January, then Wikipedia is an invisible pink unicorn" -- is true. SemanticMantis (talk) 19:22, 11 February 2013 (UTC)[reply]
All very interesting thanks for your honest input. Does Empty space exist?165.212.189.187 (talk) 19:36, 11 February 2013 (UTC)[reply]
Yes, at least conceptually. So you can make meaningful statements about it. --Stephan Schulz (talk) 19:45, 11 February 2013 (UTC)[reply]
Interesting, I thought the current most correct answer to the question "does empty space exist?" was NO. Vespine (talk) 21:46, 11 February 2013 (UTC)[reply]
I saw a documentary a while ago about post-glacial rebound. They interviewed an old woman, who had lived her whole life in the same place (it was either in northern Sweden or in Finland). They went to the house she had lived in as a child; she said that then the family rowing boat was tied up right outside the window. Now the ground had risen such that the water's edge was about 10 metres (horizontally) from the house. This paper gives the rate of rebound in Skellefteå at nearly 1cm/year; given that there was maybe 80 years between the woman's early recollection and the documentary being made, that's consistent with the land on which her house stood being maybe 80cm higher than when she was a little girl. This is a geological (and not just a local) phenomenon, as all of Britain and Fennoscandia is tilting in this manner (the woman just lived somewhere where the effect was particularly high, and particularly evident). So that's not just recorded history, that's living memory. -- Finlay McWalterჷTalk 17:30, 11 February 2013 (UTC)[reply]
I don't think so. Volcanoes, for example, have only limited effect in a limited region without significant and large-scale stratigraphcal change. In that sense, mining activities by humans will also count as geological change. --PlanetEditor (talk) 18:01, 11 February 2013 (UTC)[reply]
Geological change happens at all scales from planetary to atomic. Sean.hoyland - talk 18:09, 11 February 2013 (UTC)[reply]
Will 1964 Alaska earthquake qualify? The earthquake was accompanied by vertical displacement over an area of about 520,000 square kilometers. .... Vertical displacements ranged from about 11.5 meters of uplift to 2.3 meters of subsidence relative to sea level. Off the southwest end of Montague Island, there was absolute vertical displacement of about 13 - 15 meters. .... This zone of subsidence covered about 285,000 square kilometers, including the north and west parts of Prince William Sound, the west part of the Chugach Mountains, most of Kenai Peninsula, and almost all the Kodiak Island group. [31]Ruslik_Zero 19:00, 11 February 2013 (UTC)[reply]
Splitting water with a vacuum
Suppose you have a small icy asteroid you'd like to terraform. You wrap it securely in a watertight membrane that reflects infrared light, you put a large tinfoil dish at the L2 point behind it to focus the sunlight, and the chunk of ice and rock turns into a mini planet with a warm water ocean, maybe some ammonia. Great, except... no air, only some water vapor. So you fenestrate your membrane with little holes that let hydrogen seep through freely, and do anything you can to make the membrane kinetically catalyze the equilibrium 2H2+O2 <-->>>>> 2H2O and the same for ammonia (but easier for that, I should think). Logically, the hydrogen should go out through the holes and never be seen again.
Now, I can read from electrolysis of water that it takes "286 kJ of electrical energy input to dissociate each mole", i.e. 16 kJ per gram of water (MW 18). Work done by pushing water through a pressure differential should be delta P * V, i.e. for one cc (gram) of boiling water (1 atm = 100 kPa to 0 atm) it would be 100000 kg/ms^2 * 0.000001 m^3 = 0.1 kgm^2/s^2 = 0.1 J. Harrumph. If I'm right, that's amazingly disappointing!
Yet my feeling is that it should happen anyway - that there should be some equilibrium value for hydrogen pressure outside the membrane. If it were encapsulated by a second membrane, then it would build up to some miniscule value and eagerly react and reenter. But it's not - the moment one molecule of H2 gets free it should head for deep space and never be heard of again. Which is weird - it's as if I expect some additional work to be done on the water by "extracting" the lack of entropy from deep space.
Where am I going wrong here? And, yes, can we actually make comfy little planets this way? Wnt (talk) 17:31, 11 February 2013 (UTC)[reply]
Wouldn't it get colder? You are essentially allowing a gas to expand which will get colder, and eventually stop the hydrogen escaping. The sun heats it up though, so that's your energy source. Ariel. (talk) 18:56, 11 February 2013 (UTC)[reply]
Can you elaborate on your calculation? As stated it makes no sense to me. What are you calculating? Dauto (talk) 21:20, 11 February 2013 (UTC)[reply]
tube amplifers
looking for information on how to build a simple tube amplifer — Preceding unsigned comment added by 98.235.151.172 (talk) 18:12, 11 February 2013 (UTC)[reply]
D-d-d-danger! High Voltage! [32][33][34] don't come crying to us when you electrocute yourself to death ;) ---- nonsenseferret 18:27, 11 February 2013 (UTC)[reply]
Computer Coolant
What is the best kind of coolant for a open-air cooling sytem for a computer? My cooling setup has a cooling tower (Kind of like a nuke reactor) but the system is boiling off the water pretty fast and the water is not carrying off as much heat as I wanted... so is there a type of coolant that will not boil off as fast and carries more heat away than water? Andrew Wiggin (talk) 20:01, 11 February 2013 (UTC)[reply]
What kinda computer are you running that needs a cooling tower? Most have heat sinks and little electric fans, and manage fine? --Jayron32 20:03, 11 February 2013 (UTC)[reply]
If you are boiling water, no change of fluid will help. See latent heat of vaporization. You are pouring too much heat into the cooling system. You need bigger pipes and faster flow.
Submerging your PC in mineral oil works well if you have SSDs and no floppy or CD/DVD drives, but it is messy when you need to work on the computer. There are many web sites describing folks who have done this. --Guy Macon (talk) 21:13, 11 February 2013 (UTC)[reply]
I use it to learn CAD stuff on... My dad is teaching me so the more power I can milk out of it the fast stuff such a renderings go. Also... A Wiggin is my real name so... ya...speaker of the dead indeed? Andrew Wiggin (talk) 20:05, 11 February 2013 (UTC)[reply]
Lots of computer enthusiasts use mineral oil. It's cheap, colorless, odorless, has a high boiling point (if it doesn't decompose before that), and is so inert that you can submerge the entire motherboard in it with no ill effect. See [35], or just google "mineral oil cooling". --140.180.243.51 (talk) 20:28, 11 February 2013 (UTC)[reply]
But the real question is how much heat can the coolant carry away. Andrew Wiggin (talk) 20:32, 11 February 2013 (UTC)[reply]
I wonder if car coolant would work and how well. Andrew Wiggin (talk) 20:51, 11 February 2013 (UTC)[reply]
Have you tried removing the cover and pointing a 15-inch box fan right at it, on high ? I've never had a computer that overheats using that setup. Sure, the noise is annoying, but so is a liquid cooling system. StuRat (talk) 21:08, 11 February 2013 (UTC)[reply]
I would Prefer not to use fans as I also wish to over clock the system. I just don't think fans will cut it with a overclocked system. Andrew Wiggin (talk) 21:12, 11 February 2013 (UTC)[reply]
It's worth a try, isn't it ? A 15 inch box fan is many times better than a CPU fan and case fan combo. And air can blow thru small gaps which water or oil won't. Air also won't cause shorts. StuRat (talk) 21:33, 11 February 2013 (UTC)[reply]
Overclocking is ok for a hobby. But the amount of extra oomph you can get out of it will not make much difference to running most programs. I doubt you will get the time back you invest for building a custom cooling system. --Stephan Schulz (talk) 21:34, 11 February 2013 (UTC)[reply]
When it comes to CAD stuff every GHZ you can milk out of the processor can save 2 hours of rendering time. Andrew Wiggin (talk) 21:52, 11 February 2013 (UTC)[reply]
Clarification, do you really mean the water is boiling, or just evaporating quickly ? StuRat (talk) 21:39, 11 February 2013 (UTC)[reply]
In that case, your cooling system isn't working. The computer shouldn't be running hotter than 100C... Certainly not so much hotter that it is actually boiling the water. You are probably causing serious damage to the components. --Tango (talk) 21:57, 11 February 2013 (UTC)[reply]
What you should do is go join a specific PC overclocking / cooling forum like extremeoverclocking.com or hardforum.com. You might by chance come across a handful of people here who have practical experience in "extreme" PC cooling, or you could go straight to a popular forum FULL of people who do it for a hobby. Join, put some pics up, ask for advice, it sounds like you have something genuinely unique and unusual, I'm sure they'll love you there and be more then happy to help. Vespine (talk) 21:40, 11 February 2013 (UTC)[reply]
Here is the thing guys. I currently have the system over clocked to almost 6 Ghz. I am aiming for 10 Ghz if I can but with the way the cooling is going I will never make it there. It is critical to me to make it as high as I can just to prove to myself it can be done.Andrew Wiggin (talk) 21:56, 11 February 2013 (UTC)[reply]
Resolved