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August 27

Noise reduction history

What 'playback only' audio noise reduction techniques were available in 1989 in the US? IE how did they go about reducing tape hiss in those days?--88.110.232.152 04:24, 27 August 2007 (UTC)[reply]

Dolby noise reduction systems were quite common in 1989, but they are not "playback only". I'm not sure there are any reasonable techniques for reducing tape hiss on playback only. Since tape hiss occurs across a very broad frequency range you can't really reduce it without also reducing everything, signal as well as noise (I could be wrong, perhaps there are some Magical Algorithms out there I haven't heard about). Still, it's best to take noise reduction steps at the recording stage. Pfly 05:59, 27 August 2007 (UTC)[reply]

On prerecorded tapes at that time dolby B was common.87.102.45.106 11:34, 27 August 2007 (UTC)[reply]

The playback-only techniques that are commonly used depend upon having a "moment of silence" from which one can deduce the various noise sources in the system (hiss, turntable rumble, etc.). One way it is done is to take the Fourier transform of the noise and decide how much noise is in each of the frequency bins. The normal (non-silent) signal is then run through the transform and any frequency bin that has about the same energy as was present in the noise sample is just dropped (zeroed out). The signal is then inverse-transformed back into an analog waveform and voila!, some of the noise has been removed.

Other techniques look for discrete events like "clicks" and "pops"; the real sound source is muted during these events and inferences are drawn (again, often using the Fourier Transform) as to what the sound would have been had the click or pop not occurred.

Back in 1989, it would have been a possible but expensive proposition. Nowadays, you can buy reasonably-priced software for your PC or Mac that will do all of this automagically (e.g., Sound Soap).

Atlant 12:07, 27 August 2007 (UTC)[reply]

thanks you. l am aware of the modern spectral subtraction techniques such as is used in Cool Edit etc. I have a reissued CD digitally remixed in 1989, where they say on the sleeve note that they have managed to remove most of the tape hiss (caused by too low a recording level) on the original master tape. What noise reduction technique are they likely to have used? (BTW is not low pass filtering as the upper frequencies are OK) Oh yeah, and why would they keep the very low amplitude and refrain from eqing the tracks back to something acceptable?--88.110.232.152 13:53, 27 August 2007 (UTC)[reply]

Plants Water Purifacation

How plants can discern the particles of substances and elements , with what special characteristic of them?Flakture 07:42, 27 August 2007 (UTC)[reply]

As a general answer, the process is based on similar concepts in both plants and animals. Membrane transport proteins are generally sensitive to the charge and size of molecules and ions. Proteins can also be "designed" to take advantage of chemical characteristics of target molecules (ie, the propensity of oxygen to bind to iron in hemoglobin). Someguy1221 08:08, 27 August 2007 (UTC)[reply]

Proteins can also be evolved to take advantage of chemical characteristics of target molecules (ie, the propensity of oxygen to bind to iron in hemoglobin).--Funnyguy555 21:49, 28 August 2007 (UTC)[reply]

Photon duality

--Aaron hart 11:30, 27 August 2007 (UTC) An Old subject but could someone please explain the current theory of partical duality. I thought I had it figured out, "Transformes from energy to matter for a brief instant durring an interaction. Durring the interaction I thought that when it existed as matter is speed was <<C then back to energy propegating through space time at C. Like the particles in a accelerator that result from a smash, these particles exist as mass and in an electromagnetic field spiral and pop into energy Likewise in space-time particles are continually popping from energy fluxuations to matter, then back to energy!., not nececerrially the pairs just similar....--Aaron hart 09:25, 27 August 2007 (UTC) ANY IDEAS From top Physicist giving up secrets that are well kept. Has it been proven that photons always travel at c, is it possible that durring an interaction velcity is <C since it interacts and transforms into mass, then imidiatly back??? Just wondering I mean is anyone really sure at this time??--Aaron hart 10:18, 27 August 2007 (UTC)[reply]

You have a very distorted idea of what matter and energy are. The interactions you think you are talking about, transformation from matter to energy, is actually not this at all. The interactions you speak of are transformations from waves to particles. Light never slows down below C when it interacts with something, and it never stops being energy (the "m" in E=mc^2 is not matter, but mass. A lot of people think that mass and energy are two very different things that can simply be interconverted, but their distinction is actually quite non-apparent in high-energy physics. A photon of "pure energy" has a measurable mass, that given by E=mc^2, and an atom holds within it the energy calculated from E=mc^2, using its measured mass. If you remove through chemical interaction some measurable amount of energy from an atom, it actually weighs that much less as well, even though you removed no particles from it). Now, what it means when light turns from a wave into a particle is that it goes from having no definite position to having a definite position (it's a very difficult concept to wrap your mind around that an object can actually not have a position. The wave the photon is said to be is a sort of map of the probability that you will find it at any particular point at a given instant in time. The very important idea to keep in mind is that before the photon interacts, it isn't actually anywhere! It has no position before it interacts, merely a probability of being found at each point in space). Now, as to what's going on in those particle accelerators is that a pair of particles, typically those generally associated with matter (ie, protons) are smashed together, and interactions that can be described only by quantum mechanics (these interactions are impossible to understand, no point trying to explain them) transform the proton into new particles. As for the particles popping in and out of existence in space time, there are two sources for these. Some are photons or or something else transforming into another type of particle, and then back. Others just, well, pop out of nowhere. These are called virtual particles. They don't actually require any energy to pop into existence, they just do. You can't detect them unless you supply them with energy somehow, but that's another story. But back to the energy/mass/matter thing. Matter is typically used to describe "that which is tangible" (atoms and their constituents). Energy, well, I have no way of explaining that one. You can read energy if you like. The most basic physical idea is that energy is a property of all particles that is conserved (the total amount of it in the universe never changes) in all possible detectable interactions, but is not a measurable physical property like mass. And mass, of course, is merely that which has inertia. I'm sure my answer will not satisfy you, please ask more, and do consider taking introductory quantum physics when you enter college (if you haven't already). Someguy1221 10:25, 27 August 2007 (UTC)[reply]

About light dropping below C, while the waves of light can do very funny things indeed, and appear to speed up or slow down, the most basic answer is that photons cannot ever change speed. This is not something to be proven or disproven, but rather it is a mathematically necessary property of photons derivable from the fundamental laws of physics. Someguy1221 10:26, 27 August 2007 (UTC)[reply]

Yes I worded it rather incorectly I did mean waves to particles which I meant to refer to as mass. Now my question is how can a photon have mass when only massless waves can travel at the speed of light? Yes I did mean that the waves convert to particles of a mass with the equivalent energy. I also knew that when chemicals interact exothermically they lose the mass equivalent to their evergy, Is this not a form of mass to energy, and could it not similary be waves to energy and energy to mass, and back. could some of the mass be changed to energy, in a wavelike packet? But the part I am interested in is the mass leaving the molecules when energy leaves, is this itself not a transformation as I have proposed?!, is not energy from a exothermic reaction transmitted as a photon? Also The particles poping into existance being photons does this not support my theory. Also I tought this was from energy fluctuations in space time......--Aaron hart 10:58, 27 August 2007 (UTC) p>s> I really appreciate someone with your calaber repling to me thoughts.[reply]

Also I was refering to energy to mass, or wave to particle the mass being equivalent to the energy of the particle. It has been a while since I have studied any of this, and quite a while since I have thought of any of it, it started to interfear with my being quote "a normal person" sorry--Aaron hart 11:30, 27 August 2007 (UTC)[reply]

I'll try to answer the new questions point by point. To say that only massless waves can travel at the speed of light is not entirely correct. The relativistic equation for kinetic energy (γ-1)m where "m" is the measurable inertial mass of the particle at rest shows that a particle with non-zero rest mass requires an infinite amount of energy to move at the speed of light, since γ = ∞ at the speed of light. This is not to say that photons have no mass, and in theory this is measurable by their gravitational influence (I can hop over to one of my labs right now and measure their momentum easily enough, but this may not satisfy you). Simply all of their mass resides in what is defined as their kinetic energy. Now, back to this mass to energy bit. Firstly, I'll declare that a particle is actually a special case of a wave. If you are concerning yourself with waves that represent a potential particle's probable position in space, then an actual particle can be thought of (ignoring a tad bit of physics at the moment) as nothing more than a Dirac delta wave. Further, one can think of energy and mass as abstract mathematical properties of waves, calculable from the use of mass/energy operators if the wave form is known. Now, an atom is a wave, or rather a superposition of many waves representing the numerous particles that can be found within. The waves have a calculable mass, and therefore energy. When what you normally think of as "pure energy" is extracted from the atom (this can be through emission of a photon or as causing something else to speed up) the wave forms will change in some way. The new wave forms, while describing the same potential particles, will have a different calculable mass/energy. Sometimes it's very easy to ascribe to some nature the type of energy one is dealing with, ie, kinetic energy, thermal energy, photonic/phononic energy, gravitational/electrical potential energy, etc. When you're dealing with a static system, like a lone atom, for example, it's very hard to ascribe the location of the energy, and we generally shrug it off and say that it's all just mass. But in truth, the "energy" is merely a calculable property of the wave functions describing that atom, and this point of view happens to work for all other forms of energy as well. This is all simply to say that classical views of the distinction between mass and energy somewhat collapse in the presence of quantum mechanics. Now, to the last question, the nature of any possible "energy fluctuations" in space-time is largely a matter of interpretation of the meaning of vacuum energy. But if you prescribe yourself to believing that particles pop into existence, then by quantum mechanics, as hard as it may be to swallow this, they didn't actually come from anything. They just start existing, and very quickly stop existing. Now unfortunately I must go to bed, but I'll be back in about 8 hours. Someguy1221 11:29, 27 August 2007 (UTC)[reply]

measurable inertial mass of the particle at rest this is where I believe it is a particle, sice it acourding to quantum mechanics it can never be at rest. Also I don't completly believe in Quantum Mechanics, Is it not based on we can not know the position and the velocity since we have to interact with it, which messes with the velocity. But this is only due to scientific measurement, not actuall fact. In fact Relitivity states the Big bang began at a Singularity, Impossible for Quantum Mechanics, but only by our measurments, thus I believe it is basied upon an unkonwn, which is unkown to observers but not God. Thus I believe in singularities, and sure it stands up to evidence, since we are observing the phenomenon, thus it fits. From our observation.--Aaron hart 11:52, 27 August 2007 (UTC) P.S. thank you[reply]

It is actually a very good question you bring up, and a common and key misunderstanding of quantum mechanics. We are often taught in high-school or introductory college-level chemistry classes that Heisenberg's uncertainty principle prohibits us from knowing the simultaneous position and momentum of a particle at a given moment in time to greater than a specified accuracy. What we normally aren't told is that it is not our inability to measure these quantities, or our inability to measure one quantity without disturbing the other (that one is only partially true. After measuring a particle's position, applying the momentum operator to the new wave function of the particle, now something like a Dirac delta, will show a very large range of probable values). Rather, the uncertainty in a particle's position/momentum arises from the fact that it doesn't actually have a single defined position and momentum. This is a very alien concept to classical physics, and why it is a common misunderstanding. As I said in above posts, the wave function, through the application of operators, gives the probability of the particle having any given position, momentum, or energy. Before the measurement is made (any arbitrary event whose outcome depends on the particle's position/momentum/energy), the particle exists only as this wave, and occupies no actual position in space. And so, from a quantum perspective, it is not that there is a limit to what we can know about the universe, but rather there is a limit to how much information can actually exist in the universe. As to the singularity thing, Quantum doesn't forbid the existence of a singularity, it merely can't explain it. But neither can relativity. Someguy1221 19:22, 27 August 2007 (UTC)[reply]

I misunderstood, I thought that a particles position and momentum could not both be measured at the same time, and that this was a conciquece of the observation. Also I don't understand how something with mass can travel at the speed or light, You site rest mass, but this is the total energy of the particle, which can not exist at rest. I just don't seem to get the concept; a particle with rest mass traveling at the speed of light, has all of its matter or more correctly mass converted to energy (i.e. electromagnetic wave), then when it interacts I believe it is possible however short lived, to transfer from energy back to mass, to interact. Then the mass transfers back to a EM wave, due to its unstable form, (being so small). Which I think all has to do with the properties of Space-time. I am most likely way off, but it makes sence to me, and it is very hard to relate in words. Note, I did not think Physicist completely understand Photon Duality at this time; only theorys with mathmatical formulas that all fit nice and neat (for the observer)--Aaron hart 21:48, 27 August 2007 (UTC)[reply]

You're right on the very last point, quantum mechanics was only ever meant to make predictions of the outcomes of experiments, nothing more. Although this has been the nature of science since Newton, moving away from attempts at providing an antiquated sense of satisfaction in explaining natural phenomena. Just a couple more points I'll repeat and/or throw in: a particle that has non-zero rest mass cannot travel at the speed of light, as this would mathematically imply infinite kinetic energy. As for photons interacting, I think all that I can say is that one can find a time-dependent wave function that describes the interaction. I've even said before, and I'll say here, if one is actually trying to solve a QM problem, it's best to ignore why any of it works, and just accept it (darned good predictions come out of it). Someguy1221 21:48, 27 August 2007 (UTC)[reply]

Thank you for your time and patience, I learned much from you, but still wonder about the natural phenomenon; and what actually takes place, but I am over 30 so I guess I should give it up. Again thank you for your time--Aaron hart 04:32, 28 August 2007 (UTC)[reply]

(Being 30 is no excuse - there are plenty of 60 year old physicists!) The problem is that you are trying to think about what "actually takes place" - implying that there is some 'classical' event going on behind the math that you simply don't understand. It's not like that at all. The reality is that all of this weird quantum stuff is what "actually takes place". Fundamental particles are indeed simultaneously particles and waves - they really don't have a well defined position or momentum - masses and time compression really does happen when objects move fast. That is literally how the universe operates. The only reason we find that strange is because we've evolved and educated ourselves in an environment where we only see and interact with slow moving macroscopic objects. This idea of 'classical' Newtonian physics is simply not how things really are - it's merely a shorthand for producing approximate answers for questions that relate to our slow moving macroscopic lifestyle. SteveBaker 11:58, 28 August 2007 (UTC)[reply]

Yes, I do believe something actually takes place, but I don't know if anyone understands it beyond mathmatical derivations from partial differentals of the wave equation. I do not believe it is a classical event, though I believe photons have simultanious mass/momentum, it is just impossible to observe, or measure. I believe that Heizenberg basied his uncertanty principle on the interaction durring observation. While Debrolie and Schrodinger used mathmatics alone, my question is did they do this to fit the observations; "this may be an ignorant question, but I am curious." And to be honest I prefer modern physics to classical newtonian, since it is more accurate and really makes more sence, I can use the equations just fine, I just wish to know what actually takes place!. I realize this may be of no concern for finding the outcome of situations, but it would be beautiful to know how particles actually interact in space-time.--Aaron hart 14:17, 28 August 2007 (UTC)[reply]

How things 'actually' happen is something that's unknowable though - two fuzzy clouds of probability approach each other - and some short (but fuzzy) amount of time later a number of differently fuzzy probability clouds come out the other side. We can't look more deeply into it than that because whatever we prod and poke the system with to examine it changes the answer...which we couldn't measure precisely in the first place. We're beyond the point where we could make progress by observing what happens and writing it down. Nowadays we have to theorize what the larger scale effect of some quantum scale event might be if it works like this rather than like that - then think up some kind of devious experiment to show that happening. Annoyingly, these experiments require greater and greater energies - resulting in exponentially more expensive equipment. The great hope of using string theory for answering these questions suffers from this exact problem. It is beautiful, elegant and has great explanatory powers - but it's unable to predict anything that we can test - so whilst it may well be right, we can't tell whether it's true or not. In the end (if we buy into string theory), the universe appears to be the result of the mathematics of probability wave packets and the more carefully we look, the less there is to see. SteveBaker 19:15, 28 August 2007 (UTC)[reply]

Riverstone pebble tiles

Hi, under composite materials I have found the above product, which I require for my knitchen worktop. Problem is I cannot find any reference to a maker, where can I source this? MMany thanks <email address redacted>

^{Email address removed — Matt Eason ^{(Talk  Contribs)} 11:01, 27 August 2007 (UTC)}[reply]

Do you mean like this http://www.alibaba.com/catalog/10880060/River_Stone_Natural_Stone_Stone_Tile_Cobble_Pebble.html or http://www.ecvv.com/offerdetail/I1424081.html ?87.102.45.106 13:14, 27 August 2007 (UTC)[reply]

Thanks, close but I need a uk supplier any thoughts?

Try searching for "riverstone pebble tiles" and restrict your search to the uk - google allows you to do this. eg http://ww.google.co.uk/search?hl=en&q=riverstone+pebble+tiles&btnG=Search&meta=cr%3DcountryUK%7CcountryGB —Preceding unsigned comment added by 83.100.251.36 (talk) 06:31, August 28, 2007 (UTC)

True open circuit- calculation

How does one achieve a true open circuit (at RF) at the end of a piece of 50 ohm coaxial cable (or other sort of line) bearing in mind that the impedance of free space is always 377 ohms? Is it possible, or must one always have some radiation from the line end?--88.110.232.152 13:41, 27 August 2007 (UTC)[reply]

ie since : $\Gamma ={Z_{L}-Z_{S} \over Z_{L}+Z_{S}}$ , is it true that gamma can only be a maximum of (377 -50)/(377+50) = 0.7658? as opposed to the ideal of 1?

A very good question, and I'm disturbed that I can't get my thought in order to answer it myself. My intuition tells me that if you enlarge the cable (keeping its impedance constant, of course) so that the centre conductor and shield are far apart, the situation would be more close to the ideal open circuit. Then again, you know how intuitions can be. Please, can someone help out here? —Bromskloss 19:23, 27 August 2007 (UTC)[reply]

I suspect that the 377 ohms is irrelevant in this case, but I can't be certain. At a guess, it's because the sequence of impedances is 50R-open-377R and not 50R-377R, so the wave is reflected from the "open" condition at the end of the cable and never sees the 377 ohms of free space beyond.

I can more easily tackle the OP's conclusion, because a simple ETDR experiment shows that an open-circuited cable has a reflection coefficient close to 1. See Figure 11 in LeCroy App Note 016 for an example. --Heron 19:32, 27 August 2007 (UTC)[reply]

Ah! but how close? cant be closer than 0.7658--88.111.124.125 20:44, 27 August 2007 (UTC)[reply]

~~So you didn't look at the example I linked to, then.~~ The example I linked to above shows a gamma of at least 0.95, as nearly as I can judge from the waveform. --Heron 19:12, 28 August 2007 (UTC)[reply]

Yes I did look at your link. It gives the same response as it does on my oscilloscope. My Q was how the hell can it be better than the 0.7856 predicted by theory. Some other responses are starting to answer this question--88.109.90.214 22:45, 28 August 2007 (UTC)[reply]

OK to help anyone pursuing this line of inquiry, I quote from the article on coaxial cable:

The characteristic impedance in ohms (Ω) is calculated from the ratio of the inner (d) and outer (D) diameters and the dielectric constant ( $\epsilon _{r}$ ). The characteristic impedance is given by $Z_{0}={\frac {138}{\sqrt {\epsilon _{r}}}}\times \log({\frac {D}{d}})$ .

Certainly a lower source impedance would make a better o/c, but still not perfect. Is perfection possible/--88.110.52.40 19:35, 27 August 2007 (UTC)[reply]

On the topic of electromagnetic radiation from an open end of a coax it can happen, but the coupling to free space is very low, and unless the coax diameter is comparable to the wavelength the escaping radio waves will be minuscule. The reason is that the radial arrangements of fields cancel out at a distance. I will see if I can find a formula for this. Graeme Bartlett 22:39, 27 August 2007 (UTC)[reply]

So you are saying less than 10% of the incident power leaks out the end of the cable-- is that correct?--88.111.187.180 22:50, 27 August 2007 (UTC)[reply]

Much less that 10%. 0.01% would be more typical. Graeme Bartlett 06:18, 28 August 2007 (UTC)[reply]

The leakage is proportional to the area of the coax between the conductors, and to the square of the frequency. Graeme Bartlett 22:48, 28 August 2007 (UTC)[reply]

That's an excellent question! You made me think, and almost had me convinced. I don't think it's true, though. The reflection coefficient is really close to 1. Why? The formula you gave is specifically for a resistive load that terminates the cable. If you hook up a resistor to the end, that's what you get. But the 377 Ohms of impedance of free space is not a resistive load! Of course, you can't measure 377 Ohms of resistance across an open circuit. Therefore, the simple formula is not applicable. --Reuben 06:44, 28 August 2007 (UTC)[reply]

Indeed. I filled a whole notebook with uglier versions of the reflection coefficient equations (practical, empirical, and theoretical equations). There are boatloads of "special case" treatments which you can find in a good textbook on circuit or antenna theory. Nimur 16:53, 29 August 2007 (UTC)[reply]

toothpaste

does toothpaste ever expire? thanks.

The Chicago librarians who researched this question say yes. [1] --Sean 14:27, 27 August 2007 (UTC)[reply]

Based on my readings, toothpaste has no expiration. There is a harmless chemical that keep it from expiringWikiPoTechizen 09:57, 3 September 2007 (UTC)[reply]

reaction machenism

conversion of suger to 2,5-dimethlyfuran with complete reaction machanism

Check your spelling; I believe you mean 2,5-Dimethylfuran. The reference materials on that page should answer your question. --M @r ē ino 17:28, 27 August 2007 (UTC)[reply]

I looks like you might be thinking of a method with Hydroxymethylfurfural as an intermediate- that page has a lot of useful mechanistic infomation (on the production from sugar).87.102.85.15 17:32, 27 August 2007 (UTC)[reply]

Telescope Physics

I know how a telescope works, however, I am a little shady on the physics of light travel and a telescope. It's fairly simple when you talk about seeing a couple of miles away with binoculars as you are seeing what is happening right then. What I don't get is when astronomers look into space billions of light years away. I understand that light travels at approx 186,000 miles per second and so what is being seen happened billions of years ago. So the question is when you look through a telescope are you simply pin pointing an area of the sky as though it were a 2-dimensional image? So the more closely you can pin point an image, the farther away from earth you can see? If I were able to take an extremely high resolution picture of the sky would I be able to see as far away with the telecope? --Beckerj99 18:25, 27 August 2007 (UTC)[reply]

Your picture needs to be both high resolution and high sensitivity, as the images of distant objects will be very faint - but the basic answer to your question is "yes" - see Hubble Deep Field, Hubble Deep Field South and Hubble Ultra Deep Field. Gandalf61 18:40, 27 August 2007 (UTC)[reply]

Awesome, I always thought that was the case, but couldn't find an article specifying it. Thx!!

Earths magnetic pole swapping

Does anyone know thy from time to time, the earths magnetic poles swap from N to S.--Aaron hart 21:30, 27 August 2007 (UTC)[reply]

Geomagnetic reversal#Causes. Someguy1221 21:50, 27 August 2007 (UTC)[reply]

Thank you, from now on I will search before I ask, It was very interesting. It has been bothering me for years. Two thumbs up for Wikipedia..--Aaron hart 14:33, 28 August 2007 (UTC)[reply]

Time dilation at the speed of light

Does time dilation become infinite at the speed of light? If so, then does time pass for light? 69.150.209.13 22:18, 27 August 2007 (UTC)[reply]

time dilation becomes infinite for light. No time passes from the point of view of light. So it is emitted and absorbed at the same time, (in its own reference frame), even when it crosses astronomical distances between galaxies. Graeme Bartlett 22:45, 27 August 2007 (UTC)[reply]

composite material, kitchen work top made of clear resin and white pebbles

Were can I source this product?

Try your local hardware or DIY store perhaps. Or stores used by builders if they serve you. Given than you're provided zero information about where you live (I'm guessing the US but who knows) we can't help more. Nil Einne 22:40, 27 August 2007 (UTC)[reply]

Again a search engine is your friend "pebble resin tile" into any generic search engine will produce results eg http://www.google.co.uk/search?hl=en&q=pebble+resin+tile&btnG=Search&meta=cr%3DcountryUK%7CcountryGB (same poster for UK as above?) gives multiple choices - is that ok?83.100.251.36 06:42, 28 August 2007 (UTC)[reply]