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April 3

Limiting Reagent

Hello. If limiting reagent R is to be completely consumed by reacting with reactant X, is adding 10% in excess of the minimum amount of reactant X needed common industry practice? Are there any safety limitations in some reactions that may prevent the use of this procedure? Thanks in advance. --Mayfare (talk) 00:20, 3 April 2009 (UTC)[reply]

It's a common rule-of-thumb for something to try when developing a process, but I don't know if it's "common industry practice" for what actually goes into a full-scale production run. Gotta consider how expensive and/or difficuult it is to buy, separate, and dispose of that unreacted material, and the effect of that excess reagent could have on the reaction and products. DMacks (talk) 01:18, 3 April 2009 (UTC)[reply]

It also depends on your industry. (Let me preface this by stating that I'm not a chemical engineer). But I just got a very fancy new wall poster of a Modern Refinery (Crude Distillation), as designed by Mustang Engineering ("you hung a poster of WHAT on your wall?") which is a supposedly super-realistic depiction of a top-of-the-line refinery as it would be designed today. (So, chemical-engineer or not, after reading this detailed document, I'm now ... qualified to comment). The poster goes through the process control units in a fairly detailed way, and for the most part it seems more likely that they allow each chemical stage to reach partial completion, say 80% or thereabouts, before taking the refined product on to the next stage (by buoyancy separation, vacuum distillation, etc.). This leaves a sort of "unrefined sludge" in the each of the reactor stages. This unreacted reagent isn't "waste product," because those chemicals continue to cook or settle or dissolve or mix, and eventually reach reaction completion; but they go out with the next batch of product at a later time. Surely the process engineers understand the limiting reagents, (most likely, it seems, these are the catalyst concentrations), but have made a decision NOT to overstock limiting reagent. This could be for many reasons - in crude refining, economic pressures dictate volume throughput, so "reaction completion" is better measured in terms of absolute quantity of output product, rather than percentage-of-total-reagents-converted (that's more of an "efficiency" metric than a "completion" metric). Also, not all catalysts are "good" in large doses - metals can leach into the product; catalysts can perform side-reactions with impurities, etc etc. Nimur (talk) 13:20, 3 April 2009 (UTC)[reply]

Only slightly on topic because you mentioned posters and this is the science reference desk; I recommend these two biochemical pathway posters. Sifaka ^talk 20:31, 3 April 2009 (UTC)[reply]

I believe it works out that it costs a lot to get the last little bit to react, so often isn't worth the expense, especially if the unreacted portion can be reacted in a later batch, as noted above. Also, having extra "X" left over isn't necessarily any better than having extra "R" left over. StuRat (talk) 16:07, 3 April 2009 (UTC)[reply]

Nimur, that poster was not the Playboy pullout I was hoping for and it took forever to download, but still quite interesting. Why are the "sour water stripper" and "sulfur recovery unit" not connected to anything else? It upsets my sensibilities as an electrical engineer to have components that are not in circuit. SpinningSpark 01:34, 4 April 2009 (UTC)[reply]

Most likely because they are batch processes, rather than continuous like the rest of the plant. (A lot of people are talking about batches, but it is more efficient to deal with continuous processes. Outside pharmaceuticals, most processes are continuous) What is not shown on that diagram are the recycle streams. You would usually expect to get partial completion and recycle some of your product stream; that is, send it back into the process to mix with your reactant stream. This gives more time for the reaction to take place. Once you've separated your product from the unreacted reactants, you would probably send them back into the product stream as well. Overall, you're looking to lose very little of the excess reactant. 86.151.238.242 (talk) 00:26, 5 April 2009 (UTC)[reply]

What's the deal with black holes and escape velocity?

People talk about how light can't escape a black hole because its escape velocity is higher than the speed of light. If you used newtonian mechanics, that would mean that something within the event horizon moving the speed of light would not be able to escape orbit, but it would still be able to go arbitrarily far out. This isn't a remotely accurate description of a black hole. Something within event horizon can't even move towards the event horizon, let alone escape. Is the thing about escape velocity just a lie-to-children? Is escape velocity even the speed of light at the event horizon? — DanielLC 04:02, 3 April 2009 (UTC)[reply]

That the Newtonian calculation gets the correct result for the radius of the event horizon is little more than a coincidence. You have to use relativity to do anything meaningful with black holes. I'm not sure the concept of an escape velocity really makes sense at the event horizon of a black hole. I think the escape velocity arbitrarily close to the event horizon (but outside it) is arbitrarily close to the speed of light, though. --Tango (talk) 05:02, 3 April 2009 (UTC)[reply]

Yes, Tango is right. The scape velocity steadily approaches the speed of light as you get closer and closer to the horizon. The OP is right that a naive interpretation of that within newtonian mechanics leads to the false impression that light can reach out of the horizon but fails to make its way all the way to infinity and is forced back in. The correct description requires relativity. To put it as simply as possible, an observer falling into the blackhole could observe a photon moving outwards at the speed of light, but from the point of view of an observer at infinity that same photon would stand still right at the horizon never inching its way out of the blackhole. Dauto (talk) 05:31, 3 April 2009 (UTC)[reply]

Indeed. The problem is that time has stopped advancing at the event horizon (at least as far as outside observers are concerned)...this is an entirely relativistic phenomenon - and I suppose you could say that it's just a mathematical coincidence that this happens at the same distance that the classical escape velocity became equal to the speed of light. Of course, since we're talking about relativity - we have to be careful to say 'for whom' this effect is happening. For an outside observer, time has stopped on the event horizon - but from the perspective of something trying to escape the black hole (at less than light-speed), they can still travel past the event horizon - although they'll eventually get dragged back in again. The tricky part is that for them, time in the rest of the universe is progressing infinitely quickly - which will have all manner of nasty consequences in terms of the energy of objects (or even light) travelling into the black hole - which will now tend to have infinite frequency (and, I suppose, infinite energy)! SteveBaker (talk) 12:19, 3 April 2009 (UTC)[reply]

Question - "... from the perspective of something trying to escape the black hole (at less than light-speed), they can still travel past the event horizon - although they'll eventually get dragged back in again" - are you sure about that ? I thought that within the event horizon all timelike curves reduce their distance from the singularity at all times, so an object within the event horizon cannot even get closer to the event horizon, let alone get past it (in other words, I think I agree with what DanielLC said above). Gandalf61 (talk) 14:45, 3 April 2009 (UTC)[reply]

Answer - NO, SteveBaker got only half of the story right. The gravitational redshift which he talks about happens even to objects falling into the hole. They seem to slowdown to a stop as they approach the horizon from the point of view of an observer at infinity. But from the point of view of an observer falling into the hole the same object reaches the sigularity at the center of the hole after a finite amount of proper time. On the other hand, an outward moving light ray sits still on the horizon from the point of view of every observer. See Black hole#Event horizon. With respect to the infinite blueshift that SteveBaker talked about, You would have to sit on the horizon in order to see that. In other words, you would have to be a photon. Any observer falling in the hole would see a finite amount of red/blue shift. But photons always 'see' an infinite amount of blueshift from the direction towards which they are moving (and an infinite amount of redshift in the oposite direction), so that's notthing specific to the fact that the observer is on the horizon. In fact, such observer would not feel anything special as s/he crosses the horizon. Dauto (talk) 15:36, 3 April 2009 (UTC)[reply]

I'm unhappy with this talk of "observers at infinity". In special relativity the word "observer" is often treated as synonymous with "inertial reference frame", but that definition doesn't carry over into general relativity. In general relativistic usage, an "observer at infinity" can only observe things at infinity; he can't directly observe an object falling into a black hole, though he can observe light emitted by that object (if it eventually escapes to infinity). You can't say "from the point of view of an observer at infinity [the] photon would stand still right at the horizon", because that photon never gets anywhere near the observer.

I agree with Tango's response, except for the last sentence. The real trouble with defining escape velocity is that you need to choose a coordinate system to do it. If you choose Schwarzschild coordinates then the radial escape velocity (dr/dt) actually goes to zero at the event horizon—it has to, because the speed of light goes to zero and the escape velocity is bounded above by that (since light outside the horizon can always escape). If you use proper Schwarzschild radial velocity (dr/dτ, where r is the Schwarzschild r and τ is proper time) then the escape velocity is actually given by the Newtonian formula at all radii. (In fact, the equation of motion in terms of r and τ is d²r/dτ² = -GM/r², the same as the Newtonian equation.) The funny thing about this, though, is that the proper speed of light is infinity, not c—so the proper Schwarzschild escape velocity approaches c at the horizon, but that's not the speed of light! I don't know what to conclude from this except what Tango said—you shouldn't trust the Newtonian analogy any farther than you can throw it straight up. -- BenRG (talk) 20:03, 3 April 2009 (UTC)[reply]

I'm sorry I made you feel unhappy. I agree with you that my usage of an 'observer at infinity' was a little lazy, but I wouldn't be so draconian to the point of saying that observers at infinity cannot talk about their interpretation of what happens at the horizon just because they cannot directly observe the horizon. May be observer is a poorly chosen word. I'm not entirely sure about what you mean by "If you choose Schwarzschild coordinates then the radial escape velocity (dr/dt) actually goes to zero at the event horizon—it has to, because the speed of light goes to zero", but, at least to me, it sounds suspiciously like saying that a outward going photon stands still at the horizon, which is exactly what I said. But from whose point of view does the speed of light go to zero? Certaily not from the point of view of an observer falling into the blackhole. It is for that reason that I (lazily) chose to describe this point of view as the point of view of an observer at infinity. Dauto (talk) 05:53, 4 April 2009 (UTC)[reply]

If an object inside a black hole could get outside the event horizon temporarily then it could fire rockets during that time and escape permanently, which clearly isn't the case since that would contradict the definition of an event horizon. --Tango (talk) 00:25, 4 April 2009 (UTC)[reply]

Generation length and human evolution

How does the length of a generation affect the rate of human evolution, given how the mutation rate increases with the parents' ages? Neon Merlin 05:35, 3 April 2009 (UTC)[reply]

I think the average age people (in particular, women) have had children has been pretty constant throughout human existence until recently, so I don't think there has been enough variation to affect evolution. I'm not sure what would have happened if, hypothetically, the average age of having children had been different. There is more involved that just the rate of mutations, though - the rate of change in the habitat is also highly relevant, and that is beyond human control (or, at least, was until recently). --Tango (talk) 06:10, 3 April 2009 (UTC)[reply]

Indeed - it would be hard to know the answer to this because significantly longer generation times have only happened in the last dozen or two generations and this has corresponded with many other changes for humanity. In modern times, our ability to evolve is constantly being counteracted by our ability to adapt our environment to our needs. For example: in previous generations, if a group migrated into a sunnier part of the world, they would gradually evolve darker skin - nowadays, we slap on sunscreen. Hence the evolutionary pressure that would tend to kill off lighter skinned people in sunny climates has been greatly attenuated. Similarly, better diets and vitamin supplements allow darker skinned people to live in less sunny areas without becoming vitamin-D deficient - hence they are less likely to evolve lighter skins. Modern medicine and social support structures may bias our natural ability to evolve away inherited diseases. For example, in previous centuries, a genetic condition which might cause infertility could be evolved away in a single generation since infertile people would be unable to pass their flawed genes onto the next generation. With modern infertility treatments, those flawed genes may be passed on to the next generation who will also require infertility treatment. Hence, a gene that would normally vanish (or at least be kept strongly in check) may flourish.

This is not to say that evolution in humans has stopped - that's highly improbable - but the criteria for being successful in reproducing and raising our children has changed - and the causes that lead to reproductive success or failure are changing. Modern human evolution also has a lot more to do with memetics than genetics - when parents pass on their knowledge, stories, biasses, culture and other memes, our ideas evolve and change over generations. Take as an example, the present cultural 'war' between the Western and Arab worlds. This is largely a result of memes causing two distinct 'memetic species' - groups of people with different memes who (largely) no longer interchange their memes and are becoming memetically diverse. Just as two genetically diverse groups will eventually become so different that they can no longer interbeed - and must therefore be classified as different species - two groups who do not exchange ideas will become memetically diverse and become unable to talk to each other. That's evolution in action!

SteveBaker (talk) 12:09, 3 April 2009 (UTC)[reply]

I've got a question about this: With modern infertility treatments, those flawed genes may be passed on to the next generation who will also require infertility treatment. Hence, a gene that would normally vanish (or at least be kept strongly in check) may flourish. I've heard similar claims in the past, and they don't make sense to me. Surely the extra hurdle of infertility treatment will discourage some women from having children. People with the gene will be less likely to pass on their genes that those without. Thus the gene is still selected against. It may not disappear in one generation but it will be pushed out eventually. Right? On the other hand my understanding of genetics comes from NPR and a really old third edition copy of The Origin of Species - I've never had any formal training. Plasticup ^T/C 14:09, 3 April 2009 (UTC)[reply]

I believe your analysis is essentially correct. The only way infertility might become more common is if the natural genetic drift would produce more infertile people with each generation than people who failed to get successful fertility treatments would remove. StuRat (talk) 15:21, 3 April 2009 (UTC)[reply]

While we don't know definitely how the longer generation times would affect human evolution, it's interesting to note that in some other animal species, it has been shown to lengthen the average life span. Of course if you think about it, this is hardly surprising Nil Einne (talk) 20:41, 3 April 2009 (UTC)[reply]

We seem to have drifted from the original Q. The answer would be that evolution should indeed speed up as a result of higher mutation rates from longer generations of humans, if all other factors are kept constant. There are, as noted previously, factors slowing evolution down, as well, or at least changing which characteristics are selected as "successful" (for example, changing from being able to raise kids to now having as many kids as possible and taking no responsibility for them whatsoever). StuRat (talk) 15:28, 3 April 2009 (UTC)[reply]

Wouldn't the higher mutation rate be counterbalanced somewhat by the fact that it takes longer to produce each new generation? I mean, where do you get more evolution: in 20 generations of 40 year-old parents, or in 40 generations of 20 year-old parents? -GTBacchus^(talk) 22:15, 3 April 2009 (UTC)[reply]

To some extent, yes. However, I don't think a 40 year old mother has just twice the mutation rate of a 20 year old, in her ova, but more like 10 times as high a rate. For example, take a look at this graph showing how dramatically Down's Syndrome increases with the age of the mother: [1]. StuRat (talk) 16:39, 5 April 2009 (UTC)[reply]

While natural selection is hard to quantify, it is possible to talk about variability and the rate at which it accumulates through mutation. There are two things to worry about: the time between successive generations and the mutation rate of germline cell DNA (not somatic cell DNA). An appropriate metric to express the mutation rate is the average number of mutations per zygote divided by the generation time. (By per zygote I mean the number of mutations accumulated during the time between when the parent was a zygote to when that parent produces its first daughter (or son) zygote) If, on average, women are having children later in their life (see generation) then the generation time has been lengthened. The question is how much does that extra time increase the amount of mutation? I couldn't find information about age vs. mutation rate and mutation accumulation rates during each portion of the cell cycle.
Imagine a case where most of the mutation takes place during DNA replication and very little takes place during other stages. Considering that the germline cell production of mothers has mostly finished shortly after birth, if the mutation rate is low while the oocyte is "waiting around" then the total number of mutations will be similar whether it is fertilized after 16 years or 26 years. Since the oocyte's DNA makes up half of the zygote's, this means that the "number mutations per zygote / generation length" ratio will be lower when the average generation length is longer. If the rate of mutation in gametes is non-constant and increases as the parents get older (i.e. it's accelerating) then you could get the opposite case where the number of mutations is more than enough to compensate for the extra generational time. Sifaka ^talk 21:55, 3 April 2009 (UTC)[reply]

Both sexes have the potential to undergo spontanous mutation during gametogenesis, but as correctly pointed out by Sifaka, oogenesis is basically complete at the time of birth, and no new eggs are generated in adulthood. The type of mutation that occurs in advanced maternal age has nothing to do with DNA mutations (as implied by StuRat) but rather nondisjunction during meiosis -- where the chromosomes fail to segregate normally -- thus leading to a higher risk for conditions such as Down syndrome. It's actually men who are more susceptible to the generation of new mutations with age, since they continually make new sperm and thus their gametes undergo many more replicative cycles over their lifespan. --- Medical geneticist (talk) 17:57, 7 April 2009 (UTC)[reply]

Quantummechanical operater

why quantum mechanics deals only with linear operater? why not with other operater ?Supriyochowdhury (talk) 11:25, 3 April 2009 (UTC)[reply]

You can apply any operator you like to any wave function you like. However, most physical processes in elementary quantum mechanics can be described as linear operators (e.g. A⁺ and A^- for quantum state increment and decrement). If you want to model a process that is actually physically observed, you need your mathematical model to match the physical phenomena. "By dumb luck" most of the important (elementary) phenomena are well-modeled with linear operations. Nimur (talk) 12:41, 3 April 2009 (UTC)[reply]

(Actually dumb luck has nothing to do with it. Physicists constructed the entire conceptual modeling framework of quantum theory, and made it quite different from the classical mechanical description of particles and energy, specifically so that the mathematical operations could be simplified while still matching the observed physical reality. In fact, they created such a conceptual model twice, and both times came up with "equivalent" mathematical operations. Nimur (talk) 13:04, 3 April 2009 (UTC) )[reply]

This is not really accurate. All operators in quantum mechanics are exactly linear, regardless of the physical system and the precision of the model. But those operators operate on the wave function, which is defined over the phase space of the system, which has one dimension for every degree of freedom. Quantum theories are nonlinear in the more ordinary three-dimensional sense. Here's an analogy in classical physics. Say you have a system of two billiard balls, a cue ball and an eight ball. In initial state A the cue ball is about to strike the eight ball and knock it into a pocket, which is final state A'. In initial state B the cue ball strikes the eight ball but it misses the pocket—final state B'. The evolution of the system in both cases is nonlinear (if it were linear the cue ball would have to pass through the eight ball instead of striking it). However, if you think of the "time evolution operator" that takes A to A' and B to B' in the phase space, that operator is linear in the following sense: if you start with pA + qB, representing a probability p of being in state A and a probability q of being in state B, then the operator has to take that to pA' + qB'. If it did anything else, it would be saying that there's some sort of interaction between possible worlds, which would be a lot weirder than interaction between billiard balls.

Quantum mechanics does have a kind of "interaction between possible worlds", but it's limited by the fact that the time evolution operator is still linear. People have devised nonlinear variations of Schrödinger's equation with the goal of solving the measurement problem, but it's hard to make a workable theory that's nonlinear in phase space. -- BenRG (talk) 13:21, 3 April 2009 (UTC)[reply]

This paper, Relativistic models of nonlinear quantum mechanics, seems to outline the issue, and cites several other papers which have attempted nonlinear generalizations. Nimur (talk) 13:34, 3 April 2009 (UTC)[reply]

Quantum Mechanics

In analytic solution of H-atom Scrodinger equation a coordinate transform from cartesian to sherical polar coordinate because although Cartesian coordinate (x,y& z)are not suitable for this problem .I read many books but don't able to know in details about this problem.Please tell me what kind of problem I face if I don't change the coordinate from Cartesian to sherical polar coordinate.Supriyochowdhury (talk) 11:43, 3 April 2009 (UTC)[reply]

Have you tried to solve the Schrodinger equation in x,y,z coordinates? You can perfectly well set up the boundary conditions if you so desire. Your algebraic representation of the problem will become unwieldy, and you will end up carrying dozens of sines and cosines around the equations. Take a look at spherical symmetry, and think about why it is easier to write a description for a circular wave field using circular-style coordinates. Nimur (talk) 12:43, 3 April 2009 (UTC)[reply]

Also see this "applications" section of the Spherical coordinate system article. "Two important partial differential equations, Laplace's equation and the Helmholtz equation, allow a separation of variables in spherical coordinates. The angular portions of the solutions to such equations take the form of spherical harmonics." In Quantum Mechanics, you will find similar easy solutions because the spherical coordinates match the physical problem description better than a cartesian grid. Nimur (talk) 12:54, 3 April 2009 (UTC)[reply]

superconductivity

Does superconductivity happen in all materials when sufficiently cooled or only in some materials ? —Preceding unsigned comment added by Rkr1991 (talk • contribs) 13:50, 3 April 2009 (UTC)[reply]

Our article entitled superconductivity starts as follows: Superconductivity is a phenomenon occurring in certain materials. Plasticup ^T/C 14:12, 3 April 2009 (UTC)[reply]

Many materials are not even conductors, let alone superconductors, at low temperature. SpinningSpark 14:53, 3 April 2009 (UTC)[reply]

I suggest the following modified Q: "Do all conductors become superconductors at a low enough temperature ? How about semiconductors ?". StuRat (talk) 15:10, 3 April 2009 (UTC)[reply]

A more interesting question. The answer, however, remains "no". As the third paragraph of superconductivity notes, many metals (such as gold, an excellent conductor) and insufficiently doped semiconductors do not display superconductivity. — Lomn 15:42, 3 April 2009 (UTC)[reply]

The next obvious question, then, is how we know that they don't become superconductors. That is, could it be that they do at a low enough temp, but we have been unable to test them that close to absolute zero ? Or, is there some theoretical reason why they can never exhibit superconductivity ? StuRat (talk) 16:38, 3 April 2009 (UTC)[reply]

In the case of semi-conductors, they cease to conduct because all the electrons have returned to the valence band and none have sufficient energy to cross the energy band gap into the conduction band. Conduction can no more take place in a completely full band than it can in a completely empty band, in fact there is a principle that maximum conductivity occurs in a half filled band. SpinningSpark 18:09, 3 April 2009 (UTC)[reply]

My understanding from physics lectures (which I may not correctly remember being as the distance in time is now measured in centuries) is that pure crystalline silicon (and presumably also germanium and carbon) are actually perfect insulators as absolute zero is approached. SpinningSpark 16:04, 3 April 2009 (UTC)[reply]

Carbon has many different crystalline forms. Some Carbon nanotubes are metalic. Dauto (talk) 19:08, 3 April 2009 (UTC)[reply]

Fairly certain we did not have those in the 1960s. I was, of course, referring to the crystal form of carbon analogous to silicon crystals, ie diamond. SpinningSpark 00:30, 4 April 2009 (UTC)[reply]

Capsaicin and squirrels

Hi. I've got a question about my backyard bird-feeder. It's full of sunflower seeds and peanut halves, and the squirrels in the neighborhood like to get into it, and eat up way too much of the bird food. There are many technologies that have been developed to keep squirrels from doing this sort of thing, and that one that appeals to me most involves the chemical capsaicin, which as you know, is what makes peppers hot.

Evidently, birds have no sensitivity to capsaicin, but mammals do. Thus, it makes sense to treat the bird food with capsaicin, and then the squirrels won't like it.

So... I bought 15 habanero peppers - very hot. I diced them into little bitty pieces, tried to crush and break the seeds, and boiled it all in water for 10 minutes or so, then I put the water in a spray bottle, and sprayed the seeds. It didn't work.

The water is kind of hot, but not 15-habaneros hot. I can spray it on a potato chip and eat it without trauma. Then I read that capsaicin is "hydrophobic", so water was probably a bad choice of substrate. A guy in a bar who claims to be a doctor, or at least a pharmacist, told me that vinegar would absorb the capsaicin quite well. However, I don't know if vinegar might make the seeds unpalatable to birds.

So... I seek an appropriate substrate for a hot pepper spray. It has to absorb the capsaicin, it should carry the hotness to the seeds and nuts, and it shouldn't make the bird food unpalatable or unhealthy for the birds. Any ideas? -GTBacchus^(talk) 16:53, 3 April 2009 (UTC)[reply]

Why not use hot sauce, like Dave's Insanity or something similar, that has done the work for you? Livewireo (talk) 17:38, 3 April 2009 (UTC)[reply]

I think I would end up spending a lot of money if I treat all the birdseed with a commercial hot sauce. I did smear a pack of "fire sauce" from Taco Bell on the outside of the feeder one time (it's a cage-type feeder, where seeds fit through the mesh). That didn't seem to do much. -GTBacchus^(talk) 17:49, 3 April 2009 (UTC)[reply]

Oil, such as sunflower oil, should be better (and birds like fat). You're probably better using a food processor or blender to dice the peppers into tiny bits rather than cooking it, and I'd probably leave the micro-diced peppers in the oil for a few days so it fully infuses into the oil. Having it in the oil should also mean that if Mr Squirrel does eat some and decides to spit it out, the spicy oil will still stick to his mouth, giving him a lasting reminder of where not to eat. 87.115.166.150 (talk) 17:40, 3 April 2009 (UTC)[reply]

If you're thinking "but what if the birds eat one of the pepper fragments" - that's the idea (indeed, that the fragments are so small that any mouthful will contain some). Peppers have evolved to be eaten by birds (for the purposes of moving their seeds around), and the reason the birds aren't affected by the capsaicin and mammals are is that the peppers have evolved this to that end. To the bird the habanero probably just tastes like a tomato. 87.115.166.150 (talk) 17:45, 3 April 2009 (UTC)[reply]

Yeah, I'm not worried about the birds getting burned. I chose capsaicin because I know birds have no sensitivity to it.

I like the oil suggestion - of course sunflower or safflower oil would be acceptable on sunflower and safflower seeds! I might want to use something other than my spray bottle, because those oils are quite a bit more viscous than water. Maybe I would treat the seed in a big jar by pouring oil in and rolling it about... -GTBacchus^(talk) 17:49, 3 April 2009 (UTC)[reply]

At the expense of sounding a bit Delia, I'd either dip or drizzle. Better yet use a fat that's solid at room temperature but that can easily be melted. A lot of bird food is made with lard, so melt some lard, chuck in the peppers and seeds, mix, put into a container, and put into the fridge. Once it's set, put a hole through it for a string. My local petstore sells little cages that such blocks can go into. Gosh, with this curried seeds in lard diet, it'll be a wonder if your birds don't end up looking like little feathered Bernard Mannings. 87.115.166.150 (talk) 17:58, 3 April 2009 (UTC)[reply]

You could experiment a bit, vinegar, oil whatever, the birds won't really mind. 65.121.141.34 (talk) 18:19, 3 April 2009 (UTC)[reply]

The RSPB says chili powder, curry powder, and various kinds of hot sauce. These birds eat better than I do. Dog Day Today (talk) 18:23, 3 April 2009 (UTC)[reply]

I don't like the oil idea because it would get on the inside of the feeder making it hard to clean and the oil could become rancid after a while. Sifaka ^talk 23:50, 3 April 2009 (UTC)[reply]

Why not just dice the peppers and chuck 'em right in with the seeds? The squirrel may eat a few seeds, but the first time he eats a pepper he'll be out of there. Lord knows I wouldn't hang around. Plasticup ^T/C 18:40, 3 April 2009 (UTC)[reply]

The squirrels could probably avoid the diced pepper bits if they're large enough. Also, the squirrels are probably patient enough to pick through the seeds until they get ones that are not peppered up. You need to make sure each seed packs a wallop. To do this, I would just blend the peppers in the blender and mix the resulting noxious slurry in with the seeds directly using a mixing bowl and a spatula. If getting a good even coat is difficult, try adding a little water to reduce the viscosity of the pepper slurry. Adding oil will make the inside of the feeder gross, especially if it becomes rancid. You may also wish to dry the seeds out before you put them in the feeder to avoid possible rotting and dirtying the inside of the feeder with pepper juice. Also make sure your peppers are really unpalatably hot. Really hot peppers make my eyes burn just by cutting them. Try a miniprep using your mix with some edible sunflower seeds. Sifaka ^talk 23:50, 3 April 2009 (UTC)[reply]

I have a follow up question. If peppers evolved to be pleasant to birds but not to mammals, how come I - a mammal - enjoy peppers? Dauto (talk) 19:05, 3 April 2009 (UTC)[reply]

Masochism? 65.121.141.34 (talk) 19:10, 3 April 2009 (UTC)[reply]

Eating peppers triggers release of endorphins. That's not the full story, but it's part of it. The other part is that you can learn to associate intrinsically unpleasant things with pleasant things that they go together with -- so for example, even though bitterness is intrinsically unpleasant, the bitterness of a gin-and-tonic is part of its appeal. Looie496 (talk) 20:19, 3 April 2009 (UTC)[reply]

Mammals have TRPV1 channels, birds do not. So its not the case that "peppers evolved to be pleasant to birds", its simply that birds do not have the genes required to detect the noxious "heat" of capsaicin and therefore they don't avoid them as mammals might. Why do you enjoy peppers? because thermosensation is complex, and the range between pleasant warmth and noxious heat depends on a large range of factors, including experience and genetics. Moreover, at some concentrations capsaicin be be an analgesic. I can guarantee, while you may enjoy some peppers, that there will be a point on the Scoville scale that eating a paper is no longer pleasant and instead becomes noxious to you. That will not happen to birds. Rockpocket 00:09, 4 April 2009 (UTC)[reply]

I think it is likely that "peppers evolved to be pleasant to birds", but would state it more completely in this context as, peppers were selected by evolution to be relatively more pleasant to birds than to mammals. The capsaicin works for the latter part - unpleasant to mammals in large quantities - so the plants that made more capsaicin might have had an advantage because mammals, which might completely digest the seeds or otherwise not distribute them as well as birds, don't eat them in large quantities. --Scray (talk) 13:10, 4 April 2009 (UTC)[reply]

A product called Squirrel Away does the capsaicin isolation for you. Does it work? Somewhat. -hydnjo (talk) 21:59, 3 April 2009 (UTC)[reply]

I read somewhere that the powder bothers birds, not because it's hot, but because it's powder. If I could see that product in a store, I'd at least read the label, and see if it relates more information than that link. My local bird store doesn't sell it, though. -GTBacchus^(talk) 22:08, 3 April 2009 (UTC)[reply]

As a data point, my birds eat powdered food, in the form of the dregs from their feed, without complaint. --Sean 22:16, 4 April 2009 (UTC)[reply]

GTBacchus: The birds and squirrels aren't exposed to powder but to seed. The seed has indeed been exposed to the powder but it sticks to the seed and so seems a natural part of the seed, not a powdery cloud. -hydnjo (talk) 23:44, 4 April 2009 (UTC)[reply]

It seems to me that you will need a dehydrator, in order to remove the excess moisture from the peppers, to prevent them from rotting in the bird house. I agree with the suggestion of using a blender to chop them into little bits. After this, dehydrate, then mix them with the seeds. You could also try mixing them with the seeds first, then dehydrate the mix. If you don't want to use a dehydrator, an oven can also be used. StuRat (talk) 16:31, 5 April 2009 (UTC)[reply]

(Removed off-topic post by User:Dweller). SteveBaker (talk) 01:47, 8 April 2009 (UTC)[reply]

neuron state during mania

According to this web page, an increase in "brain derived neurotrophic factor" (BDNF) improves neuron health and leads to recovery from clinical depression. But what about mania? What is the state of neurons during, say, hypomania? Could it be that it is the opposite from that of depression, i.e., the neurons swell and establish many more synapses with other neurons? --Halcatalyst (talk) 18:31, 3 April 2009 (UTC)[reply]

First, the idea that BDNF is the "final common pathway" for depression is very much a minority view. Second, as a general rule of thumb, treatments that improve unipolar depression tend to make mania worse. In any case it's unlikely that any relevant effects have much to do with overall neuron health. Looie496 (talk) 18:38, 3 April 2009 (UTC)[reply]

Yes, I know that "treatments that improve unipolar depression tend to make mania worse" (athough the point I appreciated at the referenced web site is that exercise also causes an increase in BDNF, which obviously has no side effects other than, perhaps, better physical health). The question is, what effects does mania produce in neurons? Could it be that excessive BDNF is produced? Or do other molecules play a role? I'm interested in what might be going on at the cellular level during mania. --Halcatalyst (talk) 19:10, 3 April 2009 (UTC)[reply]

I beg to differ: exercise has a bunch of side effects, we just don't usually describe them that way. Anyhow, the cellular change that seems to be drawing the greatest interest at the moment in regard to mania is an overexpression of a substance called glycogen synthase kinase 3beta (GSK3B). There's also strong evidence implicating a circadian gene called CLOCK in at least some cases. But basically it's still a wide-open question. Looie496 (talk) 20:14, 3 April 2009 (UTC)[reply]

Interesting: I see that GSK3B is involved in "energy metabolism, neuronal cell development" and that CLOCK "encodes proteins regulating circadian rhythm." Makes sense: one of the first signs of mania is that sleep gets out of kilter (a problem with depression too) and of course being excessively energized or deenergized is also a defining component of the disorder(s). Thanks for your response. I'm always interested in knowing more, though I'm pushing the limits of my understanding. :) --Halcatalyst (talk) 01:05, 4 April 2009 (UTC)[reply]

Mike Phelps

I recall reading somewhere that Michael Phelps (8x gold medals in 2008 Olympics) eats 12,000 kcal a day of food. I presume its because of his workout routine. My question is, do athletes (M.P. in particular) have a higher average body temperature compared to less physically fit individuals? 65.121.141.34 (talk) 19:03, 3 April 2009 (UTC)[reply]

Mostly not. He eats that 12,000 kcal because he's doing aerobic exercise for 5 or more hours a day. During that exercise he's obviously hotter (and such a huge duration will push the average up) although of course the thermoregulation his body employs will keep even the highest temperature down pretty well, so his body temperature won't be that much hotter than normal. When he's not exercising his body temperature should be normal, again because of thermal homeostasis. Now he does burn more calories when resting than normal people, because he has a higher muscle mass, and muscle needs calories to keep alive (so the well-muscled have a higher metabolic rate than the low-muscled). 87.115.166.150 (talk) 23:46, 3 April 2009 (UTC)[reply]

Unless he's in hot water, Phelps's particular sport is going to keep his body temperature down pretty effectively though.John Z (talk) 22:11, 4 April 2009 (UTC)[reply]

I assume you meant "hot water" physically. DMacks (talk) 09:12, 5 April 2009 (UTC)[reply]

There is a fairly efficient homeostasis mechanism to control normal human body temperature. Despite this, during exercise, the core body temperature can rise to over 40°C. Fitter people are able to generate higher power output than the unfit. Fit people generate more heat and need to dissipate this more quickly. This is partially achieved by increased blood flow to the skin. Also, fit people can tolerate a higher core temperature during exercise than the unfit. [Short answer: yes.] Axl ¤ [Talk] 11:36, 6 April 2009 (UTC)[reply]

pleural tap

does this problem occur in men as well as women? —Preceding unsigned comment added by 71.185.151.107 (talk) 21:42, 3 April 2009 (UTC)[reply]

I'm not sure what problem you are referring to, as a pleural tap (or thoracentesis) is a medical procedure to treat fluid or air in the pleural space. If you are asking whether accumulation of air and/or fluid can occur in both men and women, the answer is yes. If you're asking whether thoracentesis can be performed on men, then the answer is yes again. Otherwise, you will have to clarify your question. Cyclonenim : Chat 22:22, 3 April 2009 (UTC)[reply]

Post edit conflict: Aaargh! Your answer made my mine completely redundant, so I am going to wikilink a bunch of words in yours to make up for it. Darn you for being faster with the save page button. Sifaka ^talk 22:30, 3 April 2009 (UTC)[reply]

Sorry :) Cyclonenim : Chat 22:39, 3 April 2009 (UTC)[reply]

Eating feces

Assuming that the individual who defecates does not have any kind of infection, does eating feces actually pose a health risk? Does it make any difference if person consuming it is different to the person producing it? Does it matter if they are different species (as long as the feces doesn't contain any foodstuff poisonous to the consuming species). Thanks. JackMacBlack (talk) 22:24, 3 April 2009 (UTC)[reply]

Yes, eating feces does pose a health risk because the gut flora, while normally benign, can cause serious disease when they escape from their normal confines. On example is bacterial sepsis following intestinal perforation. Before you ask, complete lack of any gut bacteria may cause other problems. Sifaka ^talk 22:37, 3 April 2009 (UTC)[reply]

We've also got an article, coprophagia, which partly addresses your question. -GTBacchus^(talk) 22:38, 3 April 2009 (UTC)[reply]

We have a whole article on diseases that can arise from this: fecal-oral route, though that article can use some work. --Bowlhover (talk) 22:08, 4 April 2009 (UTC)[reply]