Wikipedia:Reference desk/Archives/Science/2013 November 2

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November 2

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Is it easy to separate individual voices from a crowd chatter?

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In case the description is too general, I would suppose there are 20 to 100 voices in every moment.--163.125.81.186 (talk) 10:55, 2 November 2013 (UTC)[reply]

If they're all equidistant from the microphone and all talking at once in roughly the same volume, it would be very hard to do. Maybe only if you could isolate a narrow pitch range and filter out the others. Everyone knows what a murmuring crowd sounds like, but it's hard to quantify it. One trick I've heard of is when making a film with a scene requiring background crowd murmur, the extras are advised to say "rhubarb" over and over, and obviously not in sync with each other. ←Baseball Bugs What's up, Doc? carrots12:10, 2 November 2013 (UTC)[reply]
Separating mixed up voices from a crowd electronically is extremely difficult (see Source separation). You can use a highly directional microphone to record only one person at a time. You could also record the crowd with multiple microphones and maybe use software to correlate sound coming from different places in the audio field - that can work reasonably well...for example, one simple trick for removing the vocals in a stereo music recording is to eliminate any sound that appears in both channels (by subtracting one from the other). Since most recordings place the singer in the center of the stereo image and all of the instruments off to the sides, the result is a monophonic recording without the vocals.
Interestingly, humans are very good at doing this - it's called "The Cocktail party effect" - and we have an article about that. You might also read Selective auditory attention and Auditory scene analysis. In part that's because we have two ears and all of those wrinkles in the ear and can process things like the phase shift of sound as it passes through the skull...with those tools, we can accurately locate a sound in three-dimensional space and turn our heads towards it with fairly high precision. That helps to narrow down the voice that we're interested in - but even so, the auditory processing that goes on in our heads is highly sophisticated and beyond what we can currently achieve with electronics and software processing.
SteveBaker (talk) 15:07, 2 November 2013 (UTC)[reply]
Note, though, that one of the first signs of hearing loss is the inability to pick specific voices out of a crowd. However, I'm not sure if this is a result of damage to the ear itself, or the post-processing which occurs in the brain to make this determination. StuRat (talk) 15:28, 2 November 2013 (UTC)[reply]
I was almost deaf in one ear (with near-perfect hearing in the other) for about a year before I got treated for it - and I can confirm that loss of hearing in one ear makes it almost impossible to separate out even two people talking at once - I'd have to turn off TV and music if I wanted to have a reasonable conversation. That was purely an inner ear issue - so loss of post-processing capability in the brain wasn't the cause. SteveBaker (talk) 15:53, 2 November 2013 (UTC)[reply]
Hmm, maybe I should get my hearing checked out. I have always found it difficult to pick up individual voices in a louder crowd - one of the main reasons I have always disliked noisy clubs and bars. I find that unless I consciously make the effort to filter what I am listening to (which is quite tiring) my auditory focus drifts involuntarily between different conversations going on around me and I often lose half a sentence of what someone is saying. Also, I hate having a conversation with a TV on in the background, although that's probably more because I am old-fashioned and feel that the person I am having a conversation with should both get and give undivided attention. Anyway, that's a rambling way of saying thanks for the heads up - I'd always considered this within the range of "normal", and now I might actually go and get it looked at by a medical professional. Equisetum (talk | contributions) 18:10, 2 November 2013 (UTC) [reply]
I don't want to get into the realms of medical advice, but I have excellent hearing (in terms of being able to detect sounds), but still suffer from the same problem. Perhaps we should both see psychiatrists rather than audiologists. Tevildo (talk) 21:45, 2 November 2013 (UTC)[reply]
The first signs of hearing loss aren't always obvious. StuRat (talk) 05:19, 3 November 2013 (UTC) [reply]
If you have as many microphones as there are speakers, and if the microphones are scattered around so that each microphone hears an independent linear combination of the voice signals (being different distances from each speaker), then theoretically it should be possible to invert the matrix of linear combinations and obtain each of the voices as a separate signal. If you have different numbers of speakers and microphones, then singular value decomposition is needed. This is the basis of MIMO (using radio rather than sound propagation), and is why your wireless router sports more than one antenna (multipath propagation due to reflection from walls and objects in your house should ensure that the amplitude and phase at each antenna is uncorrelated even though the antennas are only of the order of a wavelength apart). --catslash (talk) 18:17, 2 November 2013 (UTC)[reply]

Are Operators in physics always linear?

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Are Operators in physics always linear?

150.203.188.105 (talk) 11:00, 2 November 2013 (UTC)[reply]

In quantum mechanics, yes. One of the postulates of quantum mechanics is "To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics" - see here. This postulate underlies quantum superposition. In classical physics there are, of course, many examples of non-linear systems - in general relativity and in fluid dynamics, for example. Gandalf61 (talk) 12:00, 2 November 2013 (UTC)[reply]
Every observable corresponds to a linear operator. You can define an infinite number of operators that don't correspond to observables, aren't Hermitian, etc. For example, squaring the wavefunction is not linear. --Bowlhover (talk) 16:45, 2 November 2013 (UTC)[reply]
No you can't do that, because that's not physical.
150.203.188.105 (talk) 18:39, 2 November 2013 (UTC)[reply]
That's like saying you can't introduce creation and annihilation operators for electrons because you can't create an electron out of the vacuum as that would violate conservation of energy and conservation of electric charge. Count Iblis (talk) 19:02, 2 November 2013 (UTC)[reply]
So are all physical operators linear?
150.203.188.105 (talk) 19:19, 2 November 2013 (UTC)[reply]
Yes, the time evolution operator is a linear operator and because any observation one can make is ultimately due to coupling the system with a measurement apparatus and letting that combined system evolve in time, observables must also be described by linear operators. Then while some have proposed non-linear time evolution as a mechanism to implement a real wave function collapse as opposed to an apparent one due to decoherence (linear collapse models are always mathematically identical to the system being coupled to additional degrees of freedom), they all suffer from violation of local energy and momentum conservation. So, there are good reasons to believe that time evolution and hence observables are exactly linear and not just to some good approximation as can be determined by experiment. Count Iblis (talk) 20:47, 2 November 2013 (UTC)[reply]
"In classical physics there are, of course, many examples of non-linear systems" – yes, and in quantum mechanics too. The linear operators in quantum mechanics operate on the phase space (the wave function). Classical mechanics is linear in phase space too, to the extent that it makes sense to talk about that. The more usual sense of nonlinearity in classical physics remains unchanged in quantum mechanics. The strong force is definitely not linear. -- BenRG (talk) 21:54, 2 November 2013 (UTC)[reply]

If energy is conserved, why do we run out of energy?

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A car converts gasoline into kinetic energy when it starts moving. Why can the kinetic energy not be converted back into gasoline when the car comes to a stop?

It's converted into heat. Next time you go for a jog in the summertime be sure to really bundle up. Really layer it on, sweater, coat, gloves, etc. Then jog around for a few kilometers then bring all that heat energy back home and use it to boil water for your dinner. (You should have saved up enough energy to bring at least a few liters of water to the boiling point, no?)
Or you could just read Regenerative brake instead. Hcobb (talk) 18:51, 2 November 2013 (UTC)[reply]
See Second law of thermodynamics. Count Iblis (talk) 18:59, 2 November 2013 (UTC)[reply]
Iblis has it, but in case the physics in that article is a little too dense, here's the "lay" version (and it's only little bit wrong, but not in any way that will screw up your understanding). The basic idea is that energy is always conserved, but entropy is not. Entropy is kinda a hard concept, but the easiest way to think of it is as energy which exists, but is unavailable to do work. As you do work, of any sort, some amount of energy you input into a system to do work is converted to heat because of friction (or some equivalent of friction in whatever system you're working in). So, energy input is always more than work output. The difference between the energy input and work output is basically entropy. The energy is there, it's just gone towards warming up the environment a bit, and not towards doing whatever useful work you were doing at the time. --Jayron32 01:28, 3 November 2013 (UTC)[reply]
This is what certain models of car actually do - they don't convert the kinetic energy in to gasoline, but into electricity which is stored in batteries for future use. The Honda Insight has a system like that. --TammyMoet (talk) 20:06, 2 November 2013 (UTC)[reply]


Certainly you can use the kinetic energy to charge up a battery - you could probably use battery power to drive a chemical reaction to turn water and CO2 back into gasoline - but the amount of chemistry involved would probably be horribly complicated. The biggest problem is that driving a chemical reaction 'backwards' will be highly inefficient - so you might be collecting electricity for weeks to make a thimble-full of gasoline - and you might also need a trunk-full of chemical plant to do the work. Basically, it's theoretically possible - but in practice, no - you can't do that. What you can do (and some cars actually do do) is to store electricity from regenerative braking - and use that to accelerate the car for the first few yards of acceleration when you pull away from a standing stop. SteveBaker (talk) 03:04, 3 November 2013 (UTC)[reply]
Also, even if you could convert kinetic energy back to gasoline with 100% efficiency, you'd still need to deal with kinetic energy lost due to friction, including air resistance. StuRat (talk) 05:16, 3 November 2013 (UTC)[reply]
You can't convert it back to the gasoline that was used to accelerate the car with 100% efficiency even in theory (i.e. assuming zero friction during acceleration and some ideal braking method that converts all of the kinetic energy back to work with 100% efficiency). The reason is that that in the ideal limit of zero friction a hypothetical car engine will still not be able to use all of the Gibbs energy difference between the gasoline and the combustion products when the latter are dumped into the atmosphere.
The reverse process of bringing back the gasoline would require at least that amount of work that you could have maximally have extracted from the gasoline, which involves processes that an ideal car engine is not going to make use of. The reason is that the car engine is required to be a device that fits into the car, you will have imposed some restrictions on the size of the car and you require the engine to deliver some finite amount of power. The theoretical limit is to be understood as the best possible efficiency that the laws of physics allows you to get under such constraints. This then implies that you can't reversibly mix the exhaust gasses with the atmosphere, what comes out of the exhaust will not have the same composition as the rest of the atmosphere. The amount of work that could have been obtained by reversibly mixing the exhaust gasses with the atmosphere is thus going to be lost.
This is not a very small amount of work (although it is an order of magnitude less than the caloric value of gasoline). E.g. given a litre of water you could in theory extract enough work to lift a weight of 10 kg about 700 meters if the ambient humidity is 60%. So, in theory you can make a water pump that will get all of its required energy from reversibly evaporating 10% of the water which will be enough to pump the rest of the water almost a kilometer up a hill. Count Iblis (talk) 16:26, 3 November 2013 (UTC)[reply]
See also Perpetual motion.-- OBSIDIANSOUL 06:52, 4 November 2013 (UTC)[reply]