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November 16

Miscellaneous questions about work and energy

1) If work is done on a system, then W = ΔE. But how can we calculate the work done by the system on its surroundings? Intuitively, I would think the answer is -ΔE, but I can't prove it :(. When I try, I run into contradictions (two planets at rest, gravitationally attracted to each other; if one planet is considered as the system, then its energy will increase. W_ext = -ΔE, then the other planet shouldn't speed up, but it does).

2) The work article says that the total work done in an isolated system is independent of the frame of reference. What's the significant/implications of this? 76.68.247.201 (talk) 02:07, 16 November 2010 (UTC)[reply]

Yes, work is antisymmetric like that: if I do work on you, you're doing negative work on me. However, your planets are each doing work on the other, but they also are losing potential energy in the process, so their actual $\Delta E=0$ . A heavy piston sliding down into a cylinder of gas (perhaps because the gas is cooling, so its pressure is dropping) does work on the gas even if it's moving at constant speed; the negative work done on it by the gas annihilates its potential energy.
It means that changing your frame of reference doesn't change, say, the laws of chemistry. If I fire a gun while on a train, the work done by the expanding gas on me and on the bullet is the same, which means that the energy of the chemical reaction is the same. --Tardis (talk) 15:53, 18 November 2010 (UTC)[reply]

It depends entirely on what conventions you are using. All that matters is that changing the direction of the work switches the sign of ΔE. Under some systems, we take the perspective of the system (chemical thermodynamics takes this perspective, thus exothermic reactions have a ΔE < 0 ). Under other systems, we take the perspective of the observer (who is part of the surroundings), thus the sign convention would be opposite. However, your intuition is correct. All other things being equal, the only difference between the direction of the work is the sign of ΔE. It becomes obvious if you place two objects on a number line. If object A pushes object B in the positive direction, then object A did +ΔE, if object B pushes object A with the same force over the same distance, then object A did -ΔE. As long as you keep the perspective on "work done on/by object A" you will always get opposite signs for those two situations. The implication of the independence of work from the frame of reference is the Law of conservation of energy. If you could vary the amount of work merely by changing the frame of reference, then total energy would not be conserved. --Jayron32 04:17, 16 November 2010 (UTC)[reply]

What you said for part 1 makes sense. For part two, why does it imply the conservation of energy? 76.68.247.201 (talk) 04:54, 16 November 2010 (UTC)[reply]

If I could change the amount of work done in an isolated system just by altering my frame of reference, that would imply that if I was in motion past an isolated system, and observed the work done in the system, that value would be different than if I was stationary relative to that system. That would imply that there were differing amounts of energy exchanged in the two situations; where would the extra energy come from or go to? --Jayron32 05:00, 16 November 2010 (UTC)[reply]

In the original example, both planets gain positive kinetic energy and the system (the two planets together) loses gravitational potential energy. Dbfirs 08:41, 16 November 2010 (UTC)[reply]

Yes, I had just realized that I forgot about PE after I posted. Thanks for the clarification! 76.68.247.201 (talk) 12:10, 16 November 2010 (UTC)[reply]

Wireless v fibre internet

Is it true that wireless internet will never attain the speeds of (optical) fibre internet due to the laws of physics? We're having a debate about an National Broadband Network in Australia at the moment and I hear this statement a lot. I'm wondering if it is theoretically true. 124.149.24.85 (talk) 10:35, 16 November 2010 (UTC)[reply]

I believe that the theorietical maximum is based on the frequency of the carrier, and light has a much higher frequency than radio. However I don't think that radio or fibre are near this maximum (though I could be wrong), so possibly tomorrow's wireless will exceed the maximum speed of today's fibre -- Q Chris (talk) 11:32, 16 November 2010 (UTC)[reply]

'Speed' can be a bit of nebulous term. It's more an issue of bandwidth. The argument goes that due to the finite range of frequencies available for wireless transmission it can only handle so much bandwidth, whereas for fibre, if you're running out of bandwidth you can just lay another optical fibre cable down and increase it. Wireless is often proposed as a solution in low population density areas for two main reasons - the wide spread of the few people makes laying the cable uneconomical, and secondly with only a small number of people accessing the wireless network each can have a bigger share of the limited bandwidth and thus achieve higher speeds than would be possible in high population density areas (but not as high as they'd get with fibre). As the saying goes, never say never, but with current technology and knowledge these limits apply. --jjron (talk) 12:13, 16 November 2010 (UTC)[reply]

Thanks. Do you know the name of the physical principle that limits the bandwidth in wireless communication?124.149.25.247 (talk) 01:25, 17 November 2010 (UTC)[reply]

You'll want to read the articles Bandwidth (signal processing) and most importantly Spectral efficiency. --Jayron32 01:35, 17 November 2010 (UTC)[reply]

It is hard to find true theoretical limits without very well defined conditions but for reasonable conditions in the next 30 years or so fibre seams to have a big advantage if you are concerned with bandwidth while wireless can theoretically have a small advantage in Round-trip delay time due to higher speed of electromagnetic waves in air than glass(300 000 km/s and approx. 200 000 km/s).

I define wireless internet as communication in the radio and microwave frequencies. (Ruling out Free-space optical communication and other high frequency devices that has very different properties than current wireless internet such as WIMAX, WLAN, 3G and 4G) The fastest wireless link I found was on 6 Gbit/s, see [1], it uses a frequency band at 85 GHz, it achieves 2.4bits/s/Hz over a distance of 250 m. Depending on definition the microwave band is up to 300 GHz wide. Depending on the signal to noise ratio the number of bits per second and Hz vary, se Spectral efficiency and Shannon–Hartley theorem. With very low noise the best modulations give about 15 bits/s/Hz, this would give a theoretical maximum capacity of 4.5 Tb/s and antenna. With MIMO several antennas (independent channels) can be combined to higher capacity, the number of antennas is theoretically unlimited but due to limits in device size, signal-processing and the need for path diversification the number of effective independent channels are probably limited to less than 10.

This gives a upper limit on the capacity of microwave communication to less than 50 Tb/s for each cell. (A device anywhere near this capacity will probably not be mobile and very directional so it has probably no place in a cellular network)

The physical bandwidth in a conventional fibre are about 150 THz (1000 nm to 2000 nm).

The fastest fibre optical link that has been demonstrated uses 10.8 THz and was able to achive 69.1 Tb/s over 240 km. So a single optical fibre can already get higher capacity than the "theoretical" upper limit for microwave communication.(http://www.ntt.co.jp/news2010/1003e/100325a.html) If they where able to use all the physical bandwidth they would get a capacity about 1 000 Tb/s. This can then be multiplied by an almost unlimited number of fibres. The fastest commercial links to date can archive 1.6 Tb/s, see Wavelength-division multiplexing.

--Gr8xoz (talk) 01:42, 17 November 2010 (UTC)[reply]

Thanks for your reply. My understanding is that the Shannon-Hartley theorem defines the maximum error free data transfer under a noisy channel. So, my guess is that the physics that puts an upper limit on wireless communication is the existence of Additive white Gaussian noise, which from my understanding models background radiation and other universal physical phenomena. Given that, I might rephrase my original question: is Shannon-Hartley theorem a fundamental physical law? Is there no way to theoretically circumvent the principle? 124.149.25.223 (talk) 02:52, 17 November 2010 (UTC)[reply]

After searching a bit, I think I have already found an answer to my own question. It's all about entropy. Thanks again to all who answered. 124.149.25.223 (talk) 03:23, 17 November 2010 (UTC)[reply]

Species identifcation for File:Rama rama.jpg

In order to expand on the image description, so the image can be moved to Commons, Is anyone on the Science Reference desk able to provide a more specific species identification?

Image is used on Fauna of Borneo if you need some indication of geographical location.

Sfan00 IMG (talk) 12:46, 16 November 2010 (UTC)[reply]

Just so you don't waste time looking it up rama rama is apparently just Malay for 'butterfly', Google yields nothing. —Preceding unsigned comment added by 86.4.183.90 (talk) 14:08, 16 November 2010 (UTC)[reply]

I can confirm that. I'll provide some more suggestions on Sfan's talk page (out of respect for the uploaders privacy). Nil Einne (talk) 18:28, 16 November 2010 (UTC)[reply]

Red v. grey squirrels

If grey squirrels had not been introduced into the UK, would red squirrels still be in such decline as they are now? In other words, have the grey squirrels merely taken up the space left by the red squirrels, or have they ousted them by direct competition or disease etc? Thanks 92.28.252.5 (talk) 13:26, 16 November 2010 (UTC)[reply]

There are various theories as to why grey squirrels have generally supplanted red squirrels - here - but the consensus does seem to be that the red squirrel has declined due to competition, rather than any other reason. Ghmyrtle (talk) 13:36, 16 November 2010 (UTC)[reply]

If you read that page carefully, you can see that in pine forests, there are about the same number of both kinds of juvenile squirrels. The reds' shorter lifespan, weaker constitutions, and less aggressive foraging behavior could easily be responsible for their decreased prevalence. Ginger Conspiracy (talk) 00:27, 17 November 2010 (UTC)[reply]

Seeing the space station

If I were to buy a medium range consumer available telescope and look at the international space station with it from earth, would I be able to see the station clearly and in detail? —Preceding unsigned comment added by 93.84.7.186 (talk) 14:11, 16 November 2010 (UTC)[reply]

The International Space Station's orbit never brings it lower than 278 km (173 miles) above ground. I'm not certain what is considered a typical "medium range consumer telescope" today, but I doubt it would have useful magnification much above 300 power. Even at 500 power, at the very best the station would look as though it was about 1/5 km (1/3 mile) away -- and usually farther, depending on where it was in its orbit and where on the ground you were. Not what you would call seeing it "clearly and in detail". --Anonymous, 14:31 UTC, November 16, 2010.

The ISS moves very fast, so any telescope would have difficulty keeping up with it.--Shantavira|^{feed me} 15:04, 16 November 2010 (UTC)[reply]

Starting about halfway down this page, there are several images (including some video) of the ISS transiting (passing in front of) the Sun and the Moon. Note that to observe a solar transit you need to have an appropriate telescope with suitable filters, and to get anything that would look space-station-like you need to have a big aperture telescope (several inches). Here's another solar transit, captured using a 160 mm (roughly 6-inch) telescope. Meanwhile, it is also possible to catch the ISS by itself — this image used a 100 mm – 3 inch – refractor.

However, what you aren't going to see when you look through the scope is the ISS hanging there in space. It's in low Earth orbit, moving at more than seventeen thousand miles an hour, and it's going to be rushing across the sky at about 1.25 degrees per second. It's very bright (when illuminated by the Sun) but very, very quick. Chasing it with your telescope is going to be difficult; you'll be lucky to see it slide briskly across your field of view. (The last link I provided above describes the challenge.) Ralf Vandebergh is an old hand at this stuff, however; he's coupled a video camera to his 10-inch Newtonian, and then stacks video frames together to get sharper, lower-noise images. With lots of time, effort, and practice, he's able to generate processed images that look like this or this. This remarkable frame may have captured an astronaut on a spacewalk. TenOfAllTrades(talk) 15:08, 16 November 2010 (UTC)[reply]

You can see the International Space Station with the naked eye; and you can photograph it directly, if you're careful. But it sounds like you want to produce an image of more than a bright dot. So let's clarify: a "medium range consumer scope" will probably not provide enough angular resolution, even in perfect weather and overflight-trajectory conditions, to image small details of the spacecraft. You will need a pretty good sized scope - we could say, eight-inch aperture as an absolute minimum, and a larger scope for better imaging. There are two optical "hard limits of physics" you need to worry about: angular resolution, and field of view. Our article angular resolution describes the physical limitations of angular resolution - it depends on your telescope aperture, so you can calculate a minimum observable feature size for any give down-range distance. The ISS is going to overflying me at about 750 km downrange for the next three weeks, so my eight-inch will allow me to resolve 2.0 meter features. (That's pretty darned good!) Now here's the really hard part: the other limiting factor, field of view, which depends on your telescope focal length. You can think of this as "magnification power." The more you magnify, the smaller the area of sky you can look at. So you're going to need a really accurate computer prediction of the ISS trajectory (which you can find from NASA's website or the various enthusiast tools linked above - I use kstars on Linux). And you're going to need a really carefully aligned telescope (I'm usually pretty darned sloppy, but you should find true astronomical north with a polar scope, weight down your scope with a really good tripod and mount, and very precisely position your sights on the expected overflight path. You'll have one spot of sky - you can't move your scope fast enough to track ISS. (Military-grade fast-tracking scopes that can pan the sky as fast as an orbit-track as an "exercise to the reader"). Now, you've got to wait - depending on how perfectly aligned your clock is to the ISS (in theory, both you and NASA are synchronizing to the GPS clocks). The ISS should enter your field of view exactly on schedule, and you'll have ... about two seconds, if you're lucky. So you'll want your imaging system to capture as many photos as possible (burst mode, or video mode). And you'll want to make sure your imagers are well coupled, optically, to your tube, so that you aren't losing resolution at the film or digital imager. Finally, be certain to run some numbers for your optical system, camera's shutter speed, ISO settings, and set your exposure settings properly - you won't be very happy if you return a photo of a black sky or a washed-out dot! If you're lucky, you'll have captured ten or so photographs of the ISS overflight. With a good deal of image post-processing (to first order, an image stack to denoise, and maybe a superresolution algorithm), and you should be able to produce fine-quality photographs of ISS. As I mentioned, my 2-meter feature resolving capability means I could (theoretically) produce a decent photo of the solar panels and individual modules; but I recognize the inherent challenges in this endeavor. So far I have not been able to capture ISS - but the orbit's approaching my latitude/longitude in end of November! Nimur (talk) 18:14, 16 November 2010 (UTC)[reply]

I've been able to catch the ISS in my telescope, but low power is preferable for this task and the view only lasts for a fraction of a second! Capturing an image would be much easier by mounting the camera "piggyback" to the telescope and shooting a long exposure photograph that captures the space station's trail. ~A H 1^(TCU) 03:29, 18 November 2010 (UTC)[reply]

Differences between clone and original

What physical differences could be between human clone and its original "prototype"? —Preceding unsigned comment added by 89.77.156.31 (talk) 18:27, 16 November 2010 (UTC)[reply]

Age? Current cloning techniques can't instantly create an exact copy, just an embryo which, in time, will grow up to be a "copy". --131.188.3.20 (talk) 18:39, 16 November 2010 (UTC)[reply]

Fingerprints and obvious stuff like height, weight and various other things. Identical twins should give you a clue although given that the clone would likely be raised in a rather different environment some differences are likely to be more pronounced. (Clones and identical twins also have a relatively different genetic history.) Nil Einne (talk) 18:47, 16 November 2010 (UTC)[reply]

New somatic mutations, epigenetic differences, telomere length, chance differences during embryogenesis, differences in neuronal connections and experience-dependent synaptic plasticity. Just to name a few. --- Medical geneticist (talk) 19:17, 16 November 2010 (UTC)[reply]

Further to Nil's post, you might find Twin study interesting. Although as it states in the article, twin studies are mostly about behavioral not physical differences. Vespine (talk) 23:02, 16 November 2010 (UTC)[reply]

Absolutely, any difference you expect between identical twins you would also expect in clones. Freckles, birthmarks, that sort of thing visually. Ginger Conspiracy (talk) 06:21, 17 November 2010 (UTC)[reply]

Monozygotic twins might be even more identical than clones, because the in utero environment for twins would be very similar, whereas the environmental experience of the clone would be quite different from the original. -- Scray (talk) 19:44, 17 November 2010 (UTC)[reply]

It's also not yet possible to clone humans, so one of the differences is that the original can exist. Paul (Stansifer) 03:28, 17 November 2010 (UTC)[reply]

Schrödinger's cat

It seems to me that this is a very bad example. Who could believe the cat could be both dead and alive? AdbMonkey (talk) 22:31, 16 November 2010 (UTC)[reply]

Much of quantum mechanics is classically non-intuitive; that does not make representative examples (such as Schrödinger's cat) "bad". Note that our article describes the cat as a reductio ad absurdum intended to critique an interpretation of quantum mechanics. Note also that by the time you get to interpretations you're really into the philosophy aspects of the science: the math is the same, but the explanation for "how" varies. — Lomn 22:40, 16 November 2010 (UTC)[reply]

I thought that was sort of "the point". Vespine (talk) 23:07, 16 November 2010 (UTC)[reply]

It isn't "both dead and alive" -- the cat's wave function is comprised of two juxtaposed but completely different states, only one of which is, has been, or will ever be real. Just because there are two disjoint juxtaposed states doesn't mean that they are both as real as the other. Ginger Conspiracy (talk) 23:55, 16 November 2010 (UTC)[reply]

What do you mean when you say that only one state is real? Are you suggesting a hidden variable theory? Rckrone (talk) 02:30, 17 November 2010 (UTC)[reply]

No, only that the wave function collapses deterministically, so only the one state which eventually prevails was ever real in the first place. Ginger Conspiracy (talk) 06:09, 17 November 2010 (UTC)[reply]

Doesn't work. Not even consistent with the two-slit experiment. If there's a real answer to which slit each photon went through, you don't get an interference pattern. --Trovatore (talk) 09:43, 17 November 2010 (UTC)[reply]

The case where the wavefunction collapses into a density function representing both of its disjoint states is distinct from the case where it can't. Ginger Conspiracy (talk) 17:33, 17 November 2010 (UTC)[reply]

"Anyone who is not shocked by quantum theory has not understood it." — Niels Bohr. --Mr.98 (talk) 02:08, 17 November 2010 (UTC)[reply]

Wouldn't the cat observe itself, thereby collapsing the juxtapostion of states. Plasmic Physics (talk) 06:25, 17 November 2010 (UTC)[reply]

Certainly, but for the purposes of the thought experiment, traditionally the cat isn't considered to observe itself. When this comes up in class, often the cat is removed from the box which then contains only the radioisotope, Geiger counter, relay, solenoid, and poison capsule, which everyone has an easier time believing can't observe itself. Ginger Conspiracy (talk) 08:20, 17 November 2010 (UTC)[reply]

It's not actually a problem that the cat observes itself. The cat is in a superposition of two states: one where it is dead, and one where it has observed itself, and to itself the wave function appears to have collapsed to the state where it's alive. Rckrone (talk) 08:48, 17 November 2010 (UTC)[reply]

Personally, I agree. The difference between classical and quantum mechanics is quite specific and can be illustrated with a number of simple thought-experiments. Schrödinger's cat is not one of those experiments. There is nothing really quantum about Schrödinger's cat; the quantum prediction for anything you might measure in this situation is the same as the classical prediction (unless you believe that there's no such thing as classical randomness—but quantum mechanics is not just classical mechanics with randomness). I think it became popular because of the vivid imagery of killing an innocent cat with poison gas, not because of any pedagogical merit. -- BenRG (talk) 08:20, 17 November 2010 (UTC)[reply]

What is obervation defined as? Does it have to involve a conscious entity? In my opinion, any interaction at all counts as some degree of observation. I think that whether an observation of this definition causes a wavefunction collapse is determined by how far information of the states propagates before being smeared by quantum mechanics; and by some sort of relationship to entropy and time. Thus, I propose a scenario: two equivalent closed cells containing one rod of atoms each. One rod is a micrometer long, the other is a centimeter long rod, both are initially in a juxtaposition of states. If both cells remain closed, there is a probability that both wavefunctions will have collapsed, there should be a larger probability for the longer rod have collapsed. This is my theory. I have never studied quantum mechanics, I just read the occational popular science journal like New Scientist, so I might be completely wrong, but seriously, what is obervation defined as? Plasmic Physics (talk) 09:20, 17 November 2010 (UTC)[reply]

Observation most definitely does not have to do with consciousness. It has somewhat to do with interaction, but more to do with information. If information is transfered from the system it was "observed". You can interact with it (for example a mirror), without transferring information and then it's not "observed". I don't follow your example with the rods. Ariel. (talk) 10:35, 17 November 2010 (UTC)[reply]

I'll explain: The mere presence of the rods within the box counts as an observation, because the electromagetic fields of the constituent atoms that make up the rods themselves transfers information to the cell walls. There is no way to completely isolate the rods, information is always transfered, so there cannot exist a juxtaposition. This is conclussion is clearly not true. As such, the degree of observation must be proportional to the probability of wavefunction collapse. This is different from the norm, where observation is a true or false property. I suggest that it is impossible to describe with perfect certainty whether a juxtaposition of states has collapsed or not without actively observing as in the Schrödinger's cat thought experiment.

Back to the example. The implications of this theory are, there is a non-zero probability that the juxtaposition of states will have collapsed for both of the cells, without having opened them. The propability is, among other things, dependent on the number of quantumn particles in a juxtaposition. The probability of the longer rod having a collapsed wavefunction in a closed cell is larger than that of the shorter rod. Plasmic Physics (talk) 11:22, 17 November 2010 (UTC)[reply]

People keep missing the one of the most important aspects of the Schrodinger's Cat paradox. Science requires an uninvolved observer, in order for results to be reproducable, which is a cornerstone part of science, we need to assume that our observations of phenomena indicate that the phenomena would go on even if we weren't there. When we say that a black hole behaves some way, we assume that the black hole behaves that way even when we aren't looking. When a biologist wants to observe the behavior of a lion in the wild, he has to do so without interacting with the lion, lest his presence alter the lion's behavior. Someone analyzing the effect of some substance on living cells wears gloves to avoid contaminating the sample with oils from their skin. Science works to try to isolate the phenomenon from the observer, so we can make universal statements about the phenomenon, and so we can assume that our results are reproducable independent of the observer. Here's the kicker with quantum mechanics: Independent observation is impossible. This is not a function of technology (lacking the proper instruments), it is a function of the nature of quantum mechanics itself. We cannot observe a quantum mechanical phenomenon without interacting with the phenomenon itself. Passive observation is impossible. Thus, when we want to say "How is this electron behaving when we aren't looking", it is actually impossible to predict with accuracy, since ANY test we would do to look at the electron involves bouncing a photon off of it, which alters the nature of the electron. We actually can't say what the electron was doing before we bounced the photon off of it. It is indeterminate. This is why quantum mechanics, as unsettling as it is to most people, is actually MORE unsettling to people of a scientific worldview, since QM violates one of the fundemental principles of science itself. --Jayron32 16:17, 17 November 2010 (UTC)[reply]

Quantum zeno effect. ~A H 1^(TCU) 03:26, 18 November 2010 (UTC)[reply]

Never mind the cat — has anyone created a box? Some way to hold a bit of information that is removed from the original source, which doesn't allow a collapse of states until it is looked into? For example, a way to look at which slit a photon passes through in a double slit experiment, and store the result in a quantum computer, and then you look at the interference pattern, and only after that you decide whether to look at the quantum computer and see which slits each photon went through (retroactively erasing the interference pattern and your memory of having looked at it and seen it) or alternatively you just erase the bits unread and the interference pattern stands. Does something like that exist? Wnt (talk) 09:19, 18 November 2010 (UTC)[reply]

http://www.newscientist.com/article/dn18669-first-quantum-effects-seen-in-visible-object.html

See delayed choice quantum eraser. --Tardis (talk) 15:46, 18 November 2010 (UTC)[reply]

There's no retroactive memory erasure, or retroactive anything, in quantum mechanics. The delayed choice quantum eraser experiment is relevant here, but there is no retroactive disappearance of interference patterns in that experiment. Don't pay much attention to the name, which was chosen more for audience appeal than accuracy, as names so often are. -- BenRG (talk) 21:04, 18 November 2010 (UTC)[reply]

I'm happy to agree that there's no concept of passive measurement in quantum mechanics, though there can be passive measurements in special cases. The quantum Zeno effect illustrates that well. I don't see how Schrödinger's cat illustrates it. In fact, opening the box is a passive measurement, since the internal state has long since collapsed to a mixed state. Schrödinger can be forgiven for not knowing that, but modern presentations can't. -- BenRG (talk) 21:04, 18 November 2010 (UTC)[reply]

Ok so it has nothing to do with the torture of a cat in a box. It has to do with the fact that physicists think they may be getting accurate results but they are not taking into consideration that their presence has probably altered the result. And this is on a micro level. It actually has nothing to do with killing a cat with radiation. And even if it did, it would be wrong, because a cat cannot both be dead and alive at the same time. The cat is really an atom, or isotope or something physics- like. AdbMonkey (talk) 20:35, 18 November 2010 (UTC)[reply]

In Schrödinger's original paper the fact that the cat was a macroscopic living being was important. One could try to recycle the same thought-experiment for a different purpose, but I see no reason to do that except that one likes fictional cat-killing as a way to engage one's students. And I think that Schrödinger's original purpose has been rendered obsolete by the modern understanding of measurement in quantum mechanics.

(And physicists most certainly do take the quantum measurement process into account. It is metaphysically troublesome that the description of isolated systems looks different from the description of systems under observation, but in practice it's known how to correctly model both cases.) -- BenRG (talk) 21:04, 18 November 2010 (UTC)[reply]

Ok, I'm just a tad confused and I'm sorry to ask this, but what key concepts should you walk away from this thought-experiment knowing? AdbMonkey (talk) 21:17, 18 November 2010 (UTC)[reply]

In practice, the cat is in one state (or, in the many worlds interpretation, we are entangled with it long before we physically open the box), because real boxes interfere with only the most simpleminded of measurements. (Consider checking for the body heat of the cat, or listening for its breathing, or x-raying the box, etc.) If one could build a sufficiently well-sealed (and insulated, and electromagnetically sealed, etc.) box, then the point of the thought experiment is that (with such a fantastic box) we really could have a cat in a mixed state, and the world really is that weird! --Tardis (talk) 06:24, 20 November 2010 (UTC)[reply]

Oh, you need a complete vacuum inside the box, and the cat can't touch the sides, because of its size the cat must be impossibly black or as close to a black body radiator as possible, and the temperature must be as close to absolute zero as possible; the matter comprising the box must be least aware of the presence of the cat - inanimate objects can observe and collapse the mixed state through field interactions. The more observing particles being directly influenced by a mixed state, the higher the chances of its collapse. As a result, this concept only works for systems small in physical size. The more particles participating in the mixed state, the harder it becomes to isolate it from observers, be they boxes, scientists, or cats. Plasmic Physics (talk) 13:06, 20 November 2010 (UTC)[reply]

Anacoustic

There seems to be something like Anacoustic, but neither google nor google books nor google scholar gives me at a first glance a good explanation. (The barium sulfate article contains the word and I have no clue what the application in Anacoustic faom could be.) --Stone (talk) 22:44, 16 November 2010 (UTC)[reply]

I'm thinking it's something like this. DMacks (talk) 22:50, 16 November 2010 (UTC)[reply]

Most likely yes!--Stone (talk) 07:17, 17 November 2010 (UTC)[reply]

It's probably anechoic, aka sound-proofing. CS Miller (talk) 22:53, 16 November 2010 (UTC)[reply]

The anacoustic zone is the atmosphere above 100 miles or so where there are still enough gas molecules to produce substantial drag, but they aren't close enough to transmit sound because the probability of molecular collisions is so low. Ginger Conspiracy (talk) 00:00, 17 November 2010 (UTC)[reply]