Wikipedia:Reference desk/Archives/Science/2011 November 4

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 4

regulator genes

what is the role/significance of regulator genes in embryo development?" — Preceding unsigned comment added by 68.229.53.198 (talk) 01:10, 4 November 2011 (UTC)

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. --Jayron32 02:37, 4 November 2011 (UTC)

These are generally called regulatory genes - there are some classic well-studied systems, like the role of Hox genes in Drosophila embryo development, which should give you a good idea. Wnt (talk) 09:44, 4 November 2011 (UTC)

message to director of wikipedia

i choosed this place to clear all my doubts but these people themselfs are very confused even they dont know the differnce between a vector and saclar quantity and insteD OF giving me the accurate answers they all temselfs are fighting so then how would i decide who's answ i have to except plz solve this problem otherwise their is no use of having such sections thank ubhaskar 03:05, 4 November 2011 (UTC) — Preceding unsigned comment added by Bhaskarandpm (talk • contribs)

There is no director of Wikipedia. I will do my best to answer, but I'm just a random person like you (Wikipedia is nothing except random people.)

A scalar quantity is a number. A vector quantity is a number with a direction. Speed is a scalar, velocity is a vector. That means that speed is represented just by a number. If I say I am going 100 km/hr in an airplane, I am expressing a speed. If I say I am going 100 km/hr due north, I am expressing a velocity. --Jayron32 03:10, 4 November 2011 (UTC)

(edit conflict) And seeing that your complaint about the intelligence of the Reference Desk regulars is terribly spelled and riddled with horrific grammar.... →Στc. 03:12, 4 November 2011 (UTC)

There is almost nothing in the world that has a 100% consensus; even the fact that the earth is round and revolves around the sun is not accepted by absolutely everyone, with no exception. This means, like with any information in life: which answer you accept is ultimately up to you. You have to take some responsibility for what knowledge you trust, the only way to do that at all reliably is employ a little critical thinking, consider the reliability of the source, seek impartial corroboration and interdependently verify anything you aren't sure about. Vespine (talk) 04:14, 4 November 2011 (UTC)

sorry for writing harsh languagebhaskar 04:53, 4 November 2011 (UTC) — Preceding unsigned comment added by Bhaskarandpm (talk • contribs)

And note that whether "speed" or "velocity" is the signed quantity really is irrelevant to the answer. The important part is that you can either add them up with the signs, and get zero average, or add them up without the signs, and a get a non-zero average. And, as explained to you, both approaches have their uses. Don't get all hung up on semantics (the names of things). StuRat (talk) 05:29, 4 November 2011 (UTC)

I assume that you've read our articles on speed and velocity (to which we should have linked in our previous replies). I would regard "signed speed" as one-dimensional velocity, but the treatment does depend on the context, so we need to know how you are using the words if we are to provide better answers. In particular, the words "speed" and "average" are used in different ways by different people, and instantaneous speed is not the same as "average speed". Dbfirs 09:30, 4 November 2011 (UTC)

The term "velocity" is often used in baseball to describe the "speed" of a pitch. However, there is an implied vector, i.e. from the pitcher's hand to at least the general direction of home plate. ←Baseball Bugs ^{What's up, Doc?} carrots→ 10:05, 4 November 2011 (UTC)

I would propose the following approach: For lots of users who contribute here you can find information relating to their education on their user pages. If there are contradictory statements about a subject, you can check these out to get an idea who is more likely to be right. For example, when asking a question about physics, someone who has a master in physics is (generally) more likely to give you the correct answer than an undergrad student majoring in chemistry. Phebus333 (talk) 23:31, 4 November 2011 (UTC)

is methyl salisilate reacting with tollen's reagent

tollens reagent + methyl salysilate = — Preceding unsigned comment added by 175.157.79.54 (talk) 13:23, 4 November 2011 (UTC)

Tollen's reagent gives a positive result (silver precipitate) when in the presence of aldehydes and α-hydroxy-ketones. It will give a negative result for carboxylic acids and basic, unsubstituted ketones. So, to answer your question, you need to look at the structure of Methyl salicylate and decide if there is an aldehyde or an α-hydroxy-ketone. Otherwise, you're going to have to do your own homework. --Jayron32 14:21, 4 November 2011 (UTC)

S.I.M card

how does a subscriber identity module work on a network? — Preceding unsigned comment added by Intr199 (talk • contribs) 14:05, 4 November 2011 (UTC)

You might find the article Subscriber Identity Module interesting. --Jayron32 14:15, 4 November 2011 (UTC)

attic insulation

I had extra cellulose insulation blowing into my attic Several years ago. I think I have a roof leak. How am I supposed to go into the attic or have a contractor go into the attic to find the leak, without them stepping on the insulation compressing it and how are they supposed to see the studs to walk on. — Preceding unsigned comment added by 92.48.194.153 (talk) 15:18, 4 November 2011 (UTC)

Contact a reputable contractor who does this sort of work all the time, and comes with a stellar reputation and good recommendations. Not to be blunt, but someone who does this sort of thing all the time should be able to deal with your situation, through experience and knowledge. In other words, while you may not be able to see how to get around the problem, I would trust someone who deals with the problem all the time to know what they are doing. --Jayron32 16:51, 4 November 2011 (UTC)

I doubt if the contractor would care if they compress the insulation. As for finding the studs, he'd start at the opening to the attic, and feel around for them, then follow the ones he finds. Eventually he would be able to predict where the rest were. If he was going to spend any amount of time up there, he would likely put down some walkways over the insulation and across the studs. Yes, this would compress the insulation more, but he wouldn't care about that. StuRat (talk) 21:19, 4 November 2011 (UTC)

Waterlogged insulation is unlikely to be very effective anyway, so I would deal with the more pressing issue (the leak) first. --Colapeninsula (talk) 12:02, 7 November 2011 (UTC)

okay is there anyway I can go up in the attic and look around without compressing the insulation?

If it isn't too deep, start at the opening and brush it aside as you move toward the leak, so you can find and walk on the ceiling joists. (You'd better be in good health and agile. Wear a dust mask.) When the problem is solved, brush it back. Alternatively, call a contractor (or two) out to give you an estimate. The roofer might tell you the whole roof needs to be replaced, in which case there won't be much need to enter the attic. Jc3s5h (talk) 15:55, 7 November 2011 (UTC)

Theoretical maximum possible temperature

Hi. The minimum possible temperature of anything is absolute zero (particles stop moving). Is there a theoretical maximum possible temperature, beyond which the particles cannot move faster or further (e.g. they would exceed light speed or something)? What is that temperature? 86.182.222.189 (talk) 15:31, 4 November 2011 (UTC)

This isn't your actual question, but just by the way, particles actually do not stop moving at absolute zero. See zero-point energy. --Trovatore (talk) 06:47, 5 November 2011 (UTC)

Strangely enough, the answer is that there is a maximum possible temperature, and it is absolute zero. This is because negative absolute temperatures are possible, but they are actually hotter than infinity -- a system at a negative temperature will lose energy to a system at any positive temperature.

Even if you only consider positive temperatures, there is no upper limit. This is because temperature is defined in thermodynamics as the slope of a certain curve, and it is possible for the curve to be vertical in some systems. A vertical curve means a temperature of infinity. Looie496 (talk) 16:02, 4 November 2011 (UTC)

This is a popular question at the Ref Desk. I would be remiss if I were to fail to mention our article on absolute hot. TenOfAllTrades(talk) 16:21, 4 November 2011 (UTC)

And, every time this question comes up, I point out that the word "temperature" refers to many different, but related, properties, depending on the context. In physics, we often define temperature as "the average kinetic energy per particle." But we don't always define temperature this way. Using the kinetic energy definition, negative temperature makes no sense (particles would have an imaginary velocity? - or negative mass? - or non-classical kinetic energy? ... All these are possibilities; but in fact the answer is much simpler). Instead, in those contexts where physicists consider negative temperature, they are referring to the entropy definition of temperature, explained here. This is not an easy or intuitive version of temperature; most people outside of the field of thermodynamics do not use this definition; so try not to jump to conclusions about the meaning of "negative" temperature until you really really understand the theoretical constructs that are being used in that scenario. For starters, read our article on temperature. To mitigate the risk of sounding like a broken record, I'm going to eschew a recommendation of my favorite book on thermal physics; instead, let's go with... Kittel's thermodynamics textbook - because... after you read it, you'll see that engineers never use negative temperature. They ... just ... don't... make ... any ... sense - at least, not when you are actually measuring an actual temperature of an actual object. Only theoretical thermodynamicists use the more esoteric definitions that permit such strange phenomena as negative absolute temperature. Even better: take a thermal physics class in a physics department; and then take a thermals class in a mechanical engineering department, and draw your own conclusions! Nimur (talk) 17:10, 4 November 2011 (UTC)

No. In physics, temperature is never defined as "average kinetic energy per particle", except in a context where all you have are single particles flying around, with no rotational or vibrational modes (or at least where those are negligible). Basically the kinetic-energy version of temperature is just simply wrong, period. In some contexts it works, but conceptually it's in error. --Trovatore (talk) 06:28, 5 November 2011 (UTC)

I also seriously question the claim that engineers never use negative temperature. Engineers work on lasers, for example, and the population inversion in a lasing medium is expressible in terms of negative temperature. Never having done engineering on lasers myself, I can't say whether those engineers actually do use the terminology, but they certainly reasonably could. --Trovatore (talk) 06:46, 5 November 2011 (UTC)

The speed of light doesn't result in a maximum temperature because temperature is actually dependant on kinetic energy, not speed (if you use the kinetic energy definition rather than the weird entropy definition that Nimur talks about above). At very high temperatures, I believe you do need to use a relativistic formula for relating temperature and speed, but you can get as hot as you like by just getting closer and closer to the speed of light (since kinetic energy approaches infinity as speed approaches the speed of light). --Tango (talk) 18:04, 4 November 2011 (UTC)

Indeed. You can always increase the amount of kinetic energy of a particle (thus increasing temperature). The closer you get to the speed of light, the less increase in velocity will result from the increasing kinetic energy. But as the common/practical definition of temperature does relate to the particle`s kinetic energy and not its speed, this means you can always increase the temperature, there is no maximum. Phebus333 (talk) 23:01, 4 November 2011 (UTC)

Look, you can't define temperature in terms of kinetic energy, period. It's just wrong. It doesn't work even approximately, in even the second-simplest case (diatomic molecules). If you have one container with helium in it (monatomic) and another with air (diatomic), and they have the same kinetic energy per particle (whether you take particle to mean atom, or whether you take it to mean molecule), they will be at different temperatures, by a very substantial margin. (I think which one is hotter depends on whether you're taking the particles to be atoms or molecules). --Trovatore (talk) 19:55, 5 November 2011 (UTC)

What if you include rotational and vibrational modes in the definition of kinetic energy? Translation is not the sole form of motion, and I don't think the "average kinetic energy" definition presupposes that it is. After all, if you say that you cannot define temperature in terms of energy, you're going to have to go through all of the physics and chemistry and theromdynamics textbooks in the world and redact the sections on the Equipartition theorem. I'm not sure that you, as a sole person, have the power to declare such a long-standing principle invalid... --Jayron32 01:10, 6 November 2011 (UTC)

You have to take the kinetic energy to be only that of the center of mass motion. Then the statement is valid, unless the temperature is so extremely low that the translational motion is frozen in the quantum mechanical ground state (long before you reach this point, the gas will have become a solid). Count Iblis (talk) 02:17, 6 November 2011 (UTC)

Then, I guess I am trying to understand Trovatore's bold statement: If temperature is unrelated to energy and to motion, then what is the entire point of the equipartition theorem? The opening sentence of the article says "In classical statistical mechanics, the equipartition theorem is a general formula that relates the temperature of a system with its average energies." Trovatore's statement seems in direct contradiction of this. Now, I fully understand that in extreme conditions, quantum effects start to take over (and I also understand that quantum mechanical effects should always be valid, but the classical model of the equipartition theorem should hold for the general definition of temperature under most conditions, at least to a workable approximation). But Trovatore seems to imply that the equipartition theorem can't be applied ever; that's not right, is it?!? --Jayron32 02:27, 6 November 2011 (UTC)

I certainly never said the equipartition theorem can't be applied! My point is that the equipartition theorem is a very different thing from "average kinetic energy per particle". If you average the energy over particles, rather than over modes, you get something that's just completely wrong. --Trovatore (talk) 04:51, 6 November 2011 (UTC)

Yes, but you are free to define the kinetic energy to be the kinetic energy in only the center of mass motion. The energy in the other modes is then defined as potential energy, despite being kinetic in nature. This is the convention in some books. They would actually not consider the average energy per mode, because the equipartition theorem is typically not valid for the vibrational modes at room temperature.

Defining kinetic and potential energy in this way is not as crazy as one may think. You would not have any problems with the kinetic energy of an atom of mass m at velocity v being exactly 1/2 m v^2, despite the fact that the electrons have some finite kinetic energy in the ground state configuration of the atom. This energy is considered to be potential energy. If you consider two atoms at rest at some distance r and consider how the electrostatic interaction between the electrons and nuclei of the two atoms shifts the ground state energy, you obtain the van der Waals energy. We then define this to be "potential energy", despite it having a kinetic component in terms of the electron motion. Count Iblis (talk) 15:36, 6 November 2011 (UTC)

I'm not sure I follow that in detail, but it sounds like special pleading and fancy bookkeeping to me. Better to understand the real (statistical mechanics) definition than to try to preserve the notion that it's just kinetic energy, if you don't watch the kinetic energy in my right hand.

On another note, if the vibrational mode is not active, then why is the specific heat for a diatomic ideal gas 5/2 R? I thought it was 1/2 R for each component of motion of the molecule, 1/2 R for rotation, and 1/2 R for vibration? Is it a full R for rotation, and if so why? --Trovatore (talk) 18:59, 6 November 2011 (UTC)

It's not my favorite definition either, but some books and lecture notes define it in this way when discussing classical statistical mechanics. They then prefer to not define temperature in terms of entropy to sidestep the fact that it is infinite in classical physics. Defining ensemble averages is possible without regularizing the infinity of the number of states witin a finite volume of phase space, and you can therefore write down a formal expression for the average kinetic energy without much problems.

In case of a diatomic molecule, the single degree of freedom for vibration yields a heat capacity of R, because there are two quadractic terms in the Hamiltonian, p^2/(2m) and 1/2 m omega^2 r^2, bith yield 1/2 R by the equipartition theorem. The total of 6 degrees of freedom (3 for each atom), decomposes as 3 for translation of center of mass, 2 for rotation (there are two axes of rotation orthogonal to the bond, the axis parallel to the bond doesn't count as that would amount to a spin degree of freedom for the atoms which we didn't consider), and one for vibration.

Then in the high temperature limit, you end up with 7/2 R, which looks to be wrong, because you would expect that it should become the same as for two independent atoms so 3 R. What is going on here is that the 7/2 R can be written as 3 R + 1/2 R, where the 1/2 R is due to the potential energy term of 1/2 m omega^2 r^2. Obvously, this is only (approximately) valid as long as you have a bound state. There is also a binding energy that doesn't contribute to the heat capacity, so the average potential energy is actually of the form 1/2 k T - e as long as the harmonic approximation for the potential holds. But clearly at large enough temperatures, the average of the potential energy will become zero, and you then get the heat capacity of 3 R . Count Iblis (talk) 21:05, 6 November 2011 (UTC)

OK, so what do these texts do in a crystal lattice, where surely you can't ignore the vibrational modes for the crystal as a whole? Are those "potential energy" as well, and how do you tell? --Trovatore (talk) 21:14, 6 November 2011 (UTC)

An old lecture note I have, discusses that after introducing quantum statistical mechanics. It simply states the definitions of the micro-canonical, canonical and grand canomical ensembles for quantum systems and then that automatically defines the temperature, which is then different from the earlier given definition for a classical system. It's not how I would set up things, but this is what the Prof. did for an advanced course. It allowed him to go through the things we already knew from earlier courses rapidly but still in a self-contained way. Count Iblis (talk) 21:33, 6 November 2011 (UTC)

Interesting, thanks. So is it fair to say that the pre-quantum formulation just didn't really have a definition for temperature in crystal lattices, except "would be in equilibrium with an ideal gas at that temperature"? Or did they do some other special bookkeeping to make it fit? --Trovatore (talk) 23:10, 6 November 2011 (UTC)

The pre-quantum definition based on (kinetic) energy is still available, but it will give the wrong result when applied directly to the crystal. See the Dulong–Petit law. Count Iblis (talk) 00:51, 7 November 2011 (UTC)

The temperatures don't have to be extreme for quantum effects to be important: eg it's not until about 1000 K that a CO₂ principal bending mode starts to "unfreeze", and that temperature is not untypical for gas-molecule vibration modes. But it is fair to say that temperature will be proportional to the average energy in degrees of freedom for which equipartion applies. Jheald (talk) 02:35, 6 November 2011 (UTC)

A monoatomic gas has a kinetic energy of E=3/2nkT. A diatomic gas has E=5/2nkT (kinetic + rotational energy). Therefore, temperature is NOT the average kinetic energy per particle, E/n; it's not even proportional to E/n because the "proportionality factor" would depend on how many degrees of freedom were available to the particles. This is all assuming an ideal gas. If we take internal energy into consideration, energy would not have such a simple relationship with temperature. Also, temperature is meaningful even in situations where there's no kinetic energy at all. If you have a system of particles with a magnetic moment, for example, and the system is under a magnetic field, you can define 1/T = dS/dE and find a perfectly meaningful temperature (which might be negative, because E has an upper bound), even though the particles could be completely stationary. --140.180.36.161 (talk) 02:44, 6 November 2011 (UTC)

Can anyone expand on why absolute hot should give an ultimate nothing-smaller-than-this minimum for dS/dE ? (Neglecting set-ups like nuclear spins etc, that can achieve negative temperatures, but aren't in equipartition with the other modes of their systems.) I can see from micro black hole why this (or perhaps just a fraction more) might be the highest temperature a black hole could reach. But why should the set-up that minimises dS/dE necessarily be a black hole? The absolute hot article doesn't seem to address this. Jheald (talk) 02:04, 6 November 2011 (UTC)

I find it surprising that no-one has as of yet mentioned the Planck temperature... Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:43, 7 November 2011 (UTC)

Elephants, Cape Buffaloes and hominids

Hi, could the high intelligence of elephants and cape buffaloes be a result of coevolution with hominids? Thanks.--Richard Peterson24.7.28.186 (talk) 17:59, 4 November 2011 (UTC)

Does their intelligence actually help them evade/defend against hominids? Their size is rather helpful, but I don't think their intelligence is particularly. --Tango (talk) 18:04, 4 November 2011 (UTC)

Unlikely. Being large and quite capable mammals, they are not prey animals of early presentient hominids; and being herbivores, they would have no reason whatsoever to interact with hominids. This is not the case with later hominins which did hunt big game, but it's highly unlikely they had any significant impacts to the vast populations of the two groups. And even long before that, AFAIK the brain sizes of the ancestors of both animals were already increasing even before the rise of human predation (hominins capable of hunting big game only arose a couple of million years ago or so). In the case of elephants, it's more a case of convergence rather than coevolution believed to have partially been caused by the necessity for fine motor control of their trunks (cf. human hands and bipedalism). Being social animals (both live in large herds with complex social dynamics) may also have something to do with it.

However, the reverse (human intelligence developing further from the necessity of having to hunt big game) is a different matter. -- Obsidi♠n Soul 18:45, 4 November 2011 (UTC)

I disagree. Hominids having been hunting large game long enough to have had some impact on them. In many cases this impact is extinction (such as with mammoths). For the survivors, an instinct to avoid humans seems common (even though such a slow-moving animal walking on only two legs and lacking fangs or claws doesn't look dangerous), but perhaps a general increase in intelligence has occurred, as well. StuRat (talk) 21:14, 4 November 2011 (UTC)

Part of the reason that made me wonder about it and ask the question wass i read wounded cape buffalos ambush hunters. That's pretty smart. But I can't see wounded buffalos ambushing lions or hyenas. But a human with spear or bow and arrow, yes.Thanks again.24.7.28.186 (talk) 22:20, 4 November 2011 (UTC)

I think wounded animals will attack any pursuer, even lions. They aren't likely to win, but considering the alternative is certain death, there's very little to lose in the attempt. StuRat (talk) 22:51, 4 November 2011 (UTC)

Yes but i don't think a wounded buffalo is greatly quicker, if at all, than lions or hyenas, so i don't think it could temporarily escape, hide and ambush a lion-it either fights the lions off, or succumbs. But it's likely it could outrun a human over short distances and hide and wait for its pursuers.-Rich Peterson24.7.28.186 (talk) 01:22, 5 November 2011 (UTC)

Evolution is not magic. In long-lived iteroparous megafauna like elephants, 10,000 years is not enough to develop the intelligence they possess now, nor does it explain the development of intelligence in other animal groups as well (and it exists in varying degrees in many different groups). Remember that the last glacial period lasted around ~100,000 years (it repeats in cycles) with animals that have already evolved during the current ice age, the Quaternary glaciation (which started around ~3 million years ago and is still ongoing though global warming may put an end to it permanently). These animals have developed their adaptations long before humans became a significant threat. Mammoths and modern elephants themselves, have existed in more or less the same form we now know today since the Zanclean age of the Pliocene (4.8 million years ago), when the only hominins around were Ardipithecus and Australopithecus who both largely ate plants and fruits, stood only around 3 to 4 feet, did not even have the technology for spears, and definitely did not eat large animals (even the possibility that they may have eaten small animals is sketchy at best). The increasing encephalization quotient evolutionary trend of proboscideans in the meantime, has been ongoing since Moeritherium, which existed since the Priabonian of the Eocene epoch (around 37 million years ago), long before Hominidae arose.

While the most recent (and still ongoing) Holocene extinction event might be anthropogenic to a very large degree in recent times, most of the mass extinctions of megafauna happened right after the end of the last glacial age. It helps if you realize that the entire human population 10,000 years ago hovered at a mere 15 million. Too small to significantly affect populations to the degree it does now, it is far more likely to be caused by the abrupt change in climate. In woolly mammoths and other animals adapted to much colder conditions, the very sudden change in temperatures would have decimated them, forcing them to migrate further and further north until they ran out of land and grazing areas. Humans may have played a part in wiping out the last survivors, but it's more a coup de grâce than anything. The current flight response animals now have at the sight of humans is a mere adaptation, given enough time it might develop into something more permanent, but as it stands it wasn't and still is not enough to save them.-- Obsidi♠n Soul 22:57, 4 November 2011 (UTC)

Note that an extinction occurring at the end of the last ice age does not preclude humans from being the agent, since this event allowed humans to spread into vast new areas (like the Americas and Siberia) and come into contact with those animals, which presumably had no defenses against humans. StuRat (talk) 23:05, 4 November 2011 (UTC)

No. You seem to be confusing encephalization quotient with animal behavior. The former is hardwired and happens because of evolution. The latter is an adaptation, a precursor to evolution but is typically simply a short-term response to new threats using preexisting traits - in this case, the animals already possessed the required EQ to learn to avoid humans (as mentioned they possessed this long before human hunters). In addition to elephants and cape buffalos, bison, musk oxen, gaurs etc. have the same responses to threat even when the attacker is not human. They developed this responses against predators in general, not humans specifically. Lions (and in earlier ages - sabre-toothed cats, dire wolves, leopards, hyenas, and similar large predators) already played a large part in the development of defensive behaviors and morphology of herd animals today, including the fact that they are forced to stay in groups for protection itself (which then lead to increased intelligence and in some cases altruism). And yes, it's coevolution, specifically an evolutionary arms race. To put it simply, before human hunters, they were not stupid peaceful grazers that simply toppled over as soon as something tried to kill them; who, at the sight of humans, suddenly decided to grow a larger braincase to counteract human ingenuity. Evolution does not happen like that. Compare dodos, moas, steller's sea cows, etc. which did not have natural predators. When they first encountered humans, did they suddenly become intelligent? No. They were wiped out.

And as I said, humans did hunt a lot of species to extinction and may have caused the extinction of a lot of megafauna, but in most cases, they were species already suffering decline from the end of the last glacial age. And But we are not responsible for why elephants have high EQ's or why cape buffalos have a highly developed all-for-one defense system.-- Obsidi♠n Soul 23:46, 4 November 2011 (UTC)

Your indentation and placement confuses me, as this does not appear to be a response to my last post. StuRat (talk) 02:23, 5 November 2011 (UTC)

It is. All of my replies are in the context of the original question of the OP. Correct me if I'm wrong, but you seem to be saying that elephant and cape buffalo intelligence were caused by human hunters?-- Obsidi♠n Soul 04:42, 5 November 2011 (UTC)

You placed your reply directly after, and indented from, my post saying "extinction occurring at the end of the last ice age does not preclude humans from being the agent". If you were responding to an earlier post of mine you should place your reply there. StuRat (talk) 16:41, 5 November 2011 (UTC)

Moot point. You are saying that human hunting pressures caused the extinction of the last glacial age megafauna and thus by extension the development of intelligence among proboscideans. Both are what I replied against. Regardless of what you may have wrote, it is what you implied. I do not seem to have guessed wrong judging by your reply below.-- Obsidi♠n Soul 21:11, 6 November 2011 (UTC)

Additionally, humans migrated out of Africa long before the end of the last glacial period and were already in contact with these animals long before the mass extinction event 10,000 years ago. See Human migration.-- Obsidi♠n Soul 00:11, 5 November 2011 (UTC)

It depends on which extinctions you're talking about. In the case of smilodons, they didn't have any contact with humans until the end of the ice age, and this contact seems likely to have caused their extinction. See smilodon#Extinction. Other species, like most mammoth species, had some contact with humans before that, but not over their entire range, which would have significantly increased the pressure on them. (There were actually small populations which escaped contact with humans then, and thus survived until much later, such as the dwarf woolly mammoth of Wrangel Island.) StuRat (talk) 02:23, 5 November 2011 (UTC)

Please read my original post:

a) I freely admitted that humans caused the extinction of survivors of the end of the glacial period (not the ice age, we are still in the middle of an ice age), thus preventing their recovery as they did in previous interglacials.

b) The massive rapid decline of megafauna species within a brief period right after abrupt changes in the climate tells you one thing: humans played the part of the executioner rather than the exterminator you seem to be suggesting. These species were already in decline, humans merely tipped them over the edge of extinction. Please read the linked articles: Holocene extinction and the Holocene extinction event.

c) This, of course, does not apply to later Holocene extinctions (the so-called "asynchronous" extinctions which happened long before or after the beginning of the current interglacial) which are decidedly almost purely anthropogenic.

d) Furthermore, I do not see how this would apply to the OP's question. Homo species which hunted big game appeared relatively late, when elephants and cape buffalos and their characteristic social structures already existed. Earlier hominids were not big game hunters and existed in even fewer numbers than the modern humans which arose later.

There are numerous scholarly works tackling this. All of them carefully differentiate the Pleistocene-Holocene border mass extinction event/decline (climate-caused) with later elimination of "survivor" pockets (human-caused). See [1], [2], [3], [4], [5], [6], etc. -- Obsidi♠n Soul 04:42, 5 November 2011 (UTC)

b) I don't get your distinction between "exterminator" and "executioner". Without humans, they would have likely survived the climate change, as they had in the past. With humans, they died out. I call that a human-caused extinction. Perhaps you meant to say that they might have survived contact with humans, had the climate not also warmed at the same time. I'm not sure that they would have.

d) You seem to be relying on the fossil record to tell us when animal intelligence developed. Thus, you must base your conclusions on brain size (technically skull cavity size) and social structure alone. Neither of these is a reliable way to determine intelligence. Thus, it's entirely possible that their intelligence increased after they came under hunting pressure from humans. Do we know this for sure ? No, but neither can we dismiss the idea. When you consider the radical changes ancient humans caused in the evolution of wolves into dogs (even before intentional breeding), it's not unreasonable to suppose that we've had some effect on every species we interact with. StuRat (talk) 16:49, 5 November 2011 (UTC)

re: b - there is a great deal of difference between the two. Some of those species were already doomed, with or without human intervention. They would survive as relict populations (exactly the kind of populations as the dwarf mammoths you mentioned) in refugia (specifically, glacial refugia), where they may still struggle to exist until such a time as favorable conditions return. Even then, these populations are inherently very very fragile. Any additional stresses can doom the entire line, in this case it was human hunting. But you can not blame humans for their disappearance, they merely dealt the final blow, they did not kill nor did they have the capacity to kill the vast numbers of the animals that once existed. Nevertheless, the rapid disappearance and the already very decimated populations remaining leave little room for adaptations to be developed and there certainly is no time for them to evolve. Remember, these are megafauna, like humans they are long-lived and reproduce slowly. Concentrated hunting on small populations will almost certainly leave them little time to produce young, much less evolve.

re: d - Occam's razor. And yes, what other evidence would you have then? Even the development of human intelligence is gauged by hominid fossil skulls. We might as well be arguing about intelligent design here if you choose the more unlikelier scenario over a likely one. The greater size of the skull cavity overwhelmingly implies a greater brain size and exceptional intelligence. Given how their [proboscideans] descendants share the same trait, it is the far more likely hypothesis than to imagine an animal which somehow had a large skull cavity with no evolutionary advantage. Intelligence is a very desirable evolutionary outcome, it should not be surprising that multiple animals have acquired it with or without human assistance. It is bordering on anthropocentric arrogance in fact to suggest that there would be no other way for them to acquire such intelligence without human intervention. Other natural predators are perfectly capable of inducing far greater evolutionary pressure than humans, in far more specialized ways than human hunting.

Wolves as well, already possessed the same (if not greater) intelligence as dogs - the reason why they were so easy to domesticate into human "packs" in the first place. And this was the result of their preexisting social system rather than human breeding. Wild wolves are most assuredly not "dumber" than domestic dogs. Furthermore, the evolutionary stresses of predation is far different than the evolutionary stresses of artificial selection. Yes we had quite a dramatic effect on other species on this planet as out populations ballooned, but no, we had no hand in a trait so basal as proboscidean intelligence. You might as well start arguing that cetaceans developed their intelligence because of recent human hunting. They share about the same lifespans and the same high intelligence as terrestrial megafauna, also reproduce slowly, and were also decimated and some driven extinction by whaling and fishing activities within relatively the same timespan as the Pleistocene megafauna extinctions. I think you would agree that that idea is preposterous. -- Obsidi♠n Soul 21:11, 6 November 2011 (UTC)

Thanks everyone. I've learned quite a bit.--Rich Peterson24.7.28.186 (talk) 15:14, 5 November 2011 (UTC)

Do cooked carrots have more calories than raw carrots? Why?

I've heard that cooked carrots have more calories than raw carrots. Cooked carrots taste sweeter, so I suspect that some of the indigestible starch in the raw carrots gets broken down to digestible, tasty, calorie-endowing sugars during the cooking process.

Do carrots have more calories? If so, why?

Thank you. — Preceding unsigned comment added by 137.131.48.128 (talk) 19:10, 4 November 2011 (UTC)

This comes from an observation in many diet plans where raw carrots are 0 "points" towards a diet and cooked carrots are 1 or 2 "points". This has nothing to do with an increase in calories in the cooked carrots. It has to do with portion size. Raw carrots are served in smaller portions than cooked carrots. Similarly, a raw potato will have few points because it is just one potato. Fries will have a lot of points because a single serving of fries tends to be two or three potatoes. -- kainaw™ 19:14, 4 November 2011 (UTC)

Comparing raw/cooked carrots to raw/fried potatoes is probably not valid in this context. Fries tend to be fried in fat or oil - which I suspect (although I am not a nutritionist) is where the calories come from. Mitch Ames (talk)

No, it appears that the opposite is true. I looked at two sources: Handbook of the Nutritional Contents of Foods, an old book containing data from 1963, and the fully modern database in Nutribase 9. The data between those two sources were roughly consistent. Depending on the source, a 100g serving of raw carrots is 41 or 42 cal, and a 100g serving of boiled and drained carrots is 31 or 35 cal. Raw carrots have more calories than boiled and drained carrots. Red Act (talk) 19:35, 4 November 2011 (UTC)

Which makes sense... heating up food burns calories. To get it to have more calories, you have to add something to it, like a pound of melted butter. -- kainaw™ 19:39, 4 November 2011 (UTC)

No, it has nothing to do with heating the food to "burn up calories". That's bullshit. The reason cooked (specifically boiled and drained carrots) has more calories per gram is that in cooking they absorb water, which adds weight to the carrots but adds no calories, bringing the calories/gram values down. --Jayron32 21:10, 4 November 2011 (UTC)

According to http://caloriecount.about.com, cooked carrots have 27 calories in 78 grams, while raw carrots have 8 calories in 28 grams. That's 0.346 cal/gram for cooked and 0.286 cal/gram for raw carrots. So, cooking the carrots seems to impart 21% more calories per gram. If carrots lose 17.5% of their weight via loss of water during the cooking process, that could account for the 21% more cal/g. — Preceding unsigned comment added by 137.131.48.128 (talk) 19:40, 4 November 2011 (UTC)

This is more complicated than simply "burn it and see how much energy you get." Cooked food is (in general) more readily absorbed, so the question comes down to how much dietary energy you're getting out of it. Olestra has an almost identical energy content as other fats if you're getting a calorie count by burning, but because of the chemistry none of that energy is absorbed into the body. Celery has plenty of calories by heat, but it doesn't get absorbed by the body during digestion. There's also the glycemic index question, where more rapidly absorbed food is considered problematic for a diet, and raw vegetables are rather an extreme on that scale. SDY (talk) 19:42, 4 November 2011 (UTC)

That's if you search http://caloriecount.about.com for "carrot", and select "Carrots – Whole, Raw." But if you select just "Carrots"or "Carrot", it lists 52 calories in a 128 g serving, or .41 cal/g, consistent with both of the sources I gave. I think the important word that explains the difference might be the word "Whole". If "Whole" includes the tops (greens) of the carrots, that would add mass, without adding as many calories per gram as what the root has. Red Act (talk) 20:37, 4 November 2011 (UTC)

I concur with some thoughts expressed above:

1) Cooking should not increase calorie content. It may convert starches to sugars, but they both have the same number of calories per gram, so that doesn't matter.

2) This, of course, assume the carrots are cooked in water alone and serve naked. If you use oil in the cooking process or put a glaze on them, then calories will be increased. Of course, a dip with the raw carrots will also increase calories there.

3) Note that you may eat a lot more carrots, if cooked. Raw carrots, especially large ones, take some effort to eat. Your jaw may soon tire of the effort.

4) Note that it may take more energy to digest raw carrots, too, and that figures in to how many calories you get from them (although not into how many they contain). StuRat (talk) 21:07, 4 November 2011 (UTC)

We evolved smaller intestines because we started to eat cooked food. Trying to get enough calories from raw natural foods is almost impossible. In a BBC documentary they showed an experiment where people were given only raw foods to eat, the same sort of diet they give Chimps in a Zoo. Within a few days, everyone was feeling very hungry despite having a full stomach all the time and not long after that people were getting diarrhea. Count Iblis (talk) 21:54, 4 November 2011 (UTC)

I would think the results would depend entirely on which raw foods they ate. I could see those results with raw potatoes, but eating raw tomatoes, berries, citrus, nuts, eggs, milk, etc. should be fine, provided you don't get a disease from an infectious agent in them. StuRat (talk) 22:48, 4 November 2011 (UTC)

The unit of energy is obviously causing some problems here. A carrot cooked or uncooked provides the same calorific 'heat' value when desiccated and 'burnt' in oxygen (which is how the caloric value is determined in the lab). However, when considering carrots from a dietary point of view, then cooked carrots have a higher “net energy value” for some of the reasons already posted. Put a mouse/human/etc. in a sealed box and measure the CO2 and other metabolic rates and one can consider that in terms of Calories (or Joules) also. However this is not quite the same type of energy conversion. When one counts calories for dietary control, one is using (by now) a very old nutritional concept – which is lacking in some practicality in this application. For instance, people eating way above the correct (?) calorific in take may still feel hunger because they are not getting the right nutrition. Dietary advice thrives on pseudo-sciences and commercial gobbledygook and and over simplifications, so whether the above mentioned adjustment to suggest the higher “net energy value” is right, or even how it is derived at I don't know. Yet in essence, it is useful to know that raw carrots have a less “net energy value” than cooked. Maybe the OP will find this article suitably elucidating, as it cover some of the comments already put: [7].For human evolution, guts and cooking there is this: [8]--Aspro (talk) 23:18, 4 November 2011 (UTC)

Reflection and smoothness

Hi, I would like to ask you a question about physics. If I have an object that it's smooth to the scale of the particle, and it is not black, will there always be reflection? I mean in assuming that there is no Refraction. Exx8 (talk) —Preceding undated comment added 22:29, 4 November 2011 (UTC).

Yes, I believe so, as long as it's not 100% transparent (and nothing is). However, the reflected rays may well be scattered, so it won't reflect like a mirror. StuRat (talk) 22:44, 4 November 2011 (UTC)

If the object's surface is smooth, why would the rays scatter? Dauto (talk) 02:00, 5 November 2011 (UTC)

It can only be so smooth, and some light will pass through many layers of molecules and reflect off atoms further down. StuRat (talk) 02:06, 5 November 2011 (UTC)

See Diffuse reflection#Mechanism. Much of diffuse reflection is not due to surface roughness, but rather due to scattering below the surface. As stated in the article "A piece of highly polished white marble remains white; no amount of polishing will turn it into a mirror."--Srleffler (talk) 17:37, 7 November 2011 (UTC)

Even a perfectly transparent substance still can reflect light rays as long as the objects refractive index is different from the refractive index of air. Dauto (talk) 02:03, 5 November 2011 (UTC)

The OP said we should assume there is no refraction, which I take to mean it has the same refractive index as air (or whatever medium surrounds the object). StuRat (talk) 02:08, 5 November 2011 (UTC)

I doubt that is what he meant. If the refractive index is the same as the surrounding medium there is no reflection at the surface. All the light is transmitted or absorbed.--Srleffler (talk) 17:37, 7 November 2011 (UTC)

The major challenge is that the particles themselves are not absolutely 'smooth'. Rayleigh scattering occurs even in 'pure' materials, due to minor inhomogeneities at the molecular and atomic scales. To be clear, this is a scattering event (with a random component) rather than a specular reflection; the scattered photons won't all bounce in the same place or the same direction. TenOfAllTrades(talk) 15:50, 6 November 2011 (UTC)