Wikipedia:Reference desk/Archives/Science/2012 April 24

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April 24

Curlews?

In February and March I have seen flocks of water birds wading in the flooded rice fields between Sacramento and Yuba City California. Their silhouettes and size are best described as Curlews but to me they appeared very dark or black in plumage. My Birds of North America field guide shows no illustrations or descriptions that fit the color of the many birds I have seen. I am curious about these birds. Can you help me with a well-educated guess? — Preceding unsigned comment added by 75.19.157.83 (talk) 00:59, 24 April 2012 (UTC)

Did they have the curlews' distinctive "long, slender, downcurved bills"?--Shantavira|^{feed me} 07:35, 24 April 2012 (UTC)

Yes, they did have the curlews'long, slender, downcurved bills. There were dozens of them, apparently feeding in the shallow water-filled fields. I've seen them on several different days (cloudy and sunny) but they all appeared much darker than any of the field guide illustrations. — Preceding unsigned comment added by 75.30.69.19 (talk) 01:49, 26 April 2012 (UTC)

M notation in chemical reactions described for rate data

In looking at chemical reaction rate data, I notice that common notation for describing the reactions often includes the symbol "M", which I initially thought stood for "any molecule" - a molecule that provides an opportunity for collision but is not changed by the reaction, implying that the rate is substantually independent of M. For example: CH₄ + M → CH₃ + H + M. However it seems that the rate coefficients (Arrhenius parameters) are very strongly dependent on just what molecule the M actually is. So why is the "M" notation used? Why not give the reaction in a form stating what the colliding molecule actually is, as in CH₄ + Ar → CH₃ + H + Ar? Is it some sort of historical reason? I've looked at many reaction rate papers and chemistry books - they all just use the notation without explaining why. Ratbone124.182.165.65 (talk) 02:56, 24 April 2012 (UTC)

Usually 'M' indicates a metal atom. Plasmic Physics (talk) 03:07, 24 April 2012 (UTC)

No, that can't be it. Refeences listed on the NIST database, e.g., Baulch, Cobos, etc, (J Phys. Chem. Ref Data, Vol 21 page 411 onward) use M to signify Ar, N₂, O₂, CH_n, you name it, most of them are not metals, and most are not atoms. Keit121.215.138.9 (talk) 03:36, 24 April 2012 (UTC)

M does mean "a molecule". The idea is that you're looking at a specific reaction here, which appears to be the homolytic (radical) cleavage of a hydrogen atom from methane to make a methyl radical and a hydrogen atom. If you want to do a series of studies as to the effect of various collision targets for such a reaction, you first define the general reaction, using "M" for the target, then you would have a table or chart in the article which would define rate law as a function of the identity of "M". That's, at least, how I would read this. M is a variable, which will later in the article be defined as a specific set of molecules. That would make the most sense to me, if I am reading your question correctly. --Jayron32 04:03, 24 April 2012 (UTC)

To put it another way, chemistry has a few algebra-like "variables". M = some molecule, just as H-X typically means some halide, or CH3-R means something bound to some side chain. The point is to allow the writer to say M, X, or R = this, that, or the other thing. Wnt (talk) 05:49, 24 April 2012 (UTC)

I've seen this M notation before in the context of gas phase kinetics like this. My recollection is that M stands for any atom or molecule whose purpose in the reaction is to give or receive energy by collision. M doesn't chemically react, it's just there to bump into. If, for example, a molecule of CH₄ collides with a molecule of M, CH₄ might end up with more energy than it had before the collision. You often see such "excited" species highlighted with an asterisk, e.g. CH₄*. The process would be expressed in an equation:

CH₄ + M → CH₄* + M

This isn't a chemical reaction in the traditional sense, it's redistribution of translational, rotational and vibrational energy. The important point is CH₄ becomes vibrationally excited as a result of the collision with M. Vibrational excitation ("chemical activation") is often necessary for a chemical reaction, such as the homolytic fission of a C–H bond:

CH₄* → CH₃ + H

The extra vibrational energy CH₄ obtained from collision with M has been spent on breaking the bond. This could not happen without activation, since bond breaking requires energy (is endergonic or endothermic?).

The two processes of (i) vibrational excitation by collision with M and (ii) bond dissociation often occur simultaneously, so are expressed together:

CH₄ + M → CH₃ + H + M

I've often heard M referred to as the "bath gas", meaning something like argon that you might use as an inert gas to dilute a reactive gas like methane for kinetic studies. I believe the general placeholder M rather than a specific symbol like Ar is used because M could be either an atom of Ar or another CH₄ molecule, it doesn't matter. M signifies the role of the molecule in the reaction, rather than its chemical identity. See Lindemann mechanism for an example of the use of M. --Ben (talk) 09:54, 24 April 2012 (UTC)

I don't actually know the answer, as like Ratbone I have never seen the reason or meaning for M notation explained. It seems to just get used. Clearly, whatever atom or molecule the M actually is DOES matter, and matters a great deal. Not only does the actual atom or molecule used greatly affects the rate, it affects the shape of the temperature dependence curve. But after thinking about it, I think that the following two reasons fit:-

Firstly, to clarify what Jayron is probably trying to say, M helps define a class of reactions. For example, CH₄ + M → CH₃ + H + M. defines a class of reactions, and the reactions CH₄ + Ar → CH₃ + H + Ar and CH₄ + N₂ → CH₃ + H + N₂ are members of this class. To make classifying reactions in this way worthwhile, there must be some characteristics common to the class. The rate equation definitely is not, but the reaction order (1st, 2nd, or 3rd as may be) is, and so is the fact that the M atom or molecule (whatever it happens to be) is totally unchanged.

Secondly, the M notation allows us to distinguish between reactions that don't change one of the collidants from those that change all of them. Consider a reaction AB + BC → A + B + BC, where A, B, and C are symbols of atoms. Two different things can be happening here. To explain it I'm going to invent a new notation - I will name the atoms individually like we name people. Let's say we have atoms A_ALEX, B_BRUCE, B_BRIAN, and C_CHARLES. Firstly we can have reaction A_ALEXB_BRUCE + B_BRIANC_CHARLES → A_ALEX + B_BRUCE + B_BRIANC_CHARLES. This is a case of the collision cleaving A_ALEXB_BRUCE into two, and leaving B_BRIANC_CHARLES unaltered. This is conveniently shown as AB + M → A + B + M, M = BC. Note the "M = BC" part - it is incomplete without this. Secondly, we can have reaction A_ALEXB_BRUCE + B_BRIANC_CHARLES → A_ALEX + B_BRIAN + B_BRUCEC_CHARLES. Here, Bruce has swapped partner and all products are changed cf the reactants. A more concise notation is AB' + B"C → A + B" + B'C. You might ask why worry about such a reaction - there appears to have been the same chemical result to the first case. But B' and B" could be different isotopes (remember that many elements naturally and artificially occur as a mix of stable isotopes) - then the physical characteristics and chemical properties of the products will be different in each case, though often perhaps not greatly different.

Keit121.215.68.210 (talk) 13:16, 24 April 2012 (UTC)

The exact numerical analysis and specific rate-constants may depend on the identity of M, but an even more significant major point is that M exists, with a coefficient of 1 in the reaction at all. That is, the equation illustrates that M plays an actual role in the reaction mechanism and does appear in whatever equations you would use for rate/equilibrium rather than really being an inert solvent/carrier-gas/etc. DMacks (talk) 14:12, 24 April 2012 (UTC)

Let's clear this up once and for all.

Atkins & de Paula, Physical Chemistry, 7th edition (2002), chapter 25 (the rates of chemical reactions), page 866.

"HCl* + M → HCl + M

Here HCl* denotes a thermally excited HCl molecule and M is a body (an unreactive molecule or the wall of the container) that removes the excess energy stored in HCl...

And again, on pages 900-901 (chapter 26, the rate laws of chain reactions):

Br₂ + M → Br^· + Br^· + M

where M is either Br₂ or H₂.

Br^· + Br^· + M → Br₂ + M*

In the termination step, the third body M removes the energy of recombination.

Finally, page 922:

O + O + M → O₂ + M

where M is an arbitrary third body, such as O₂ in an 'oxygen-only' atmosphere, which helps to conserve angular momentum.

Several different molecules and the walls of the container will all act as M in any given reaction. It is not possible to say that M = the bath gas Ar or M = CH₄ or M = the wall of the container. All three are possible and therefore all three will happen.

You can say one thing about the identity of M: some species are statistically much more likely to act as M. For example, imagine a gaseous mixture of bromine, Br₂, and argon, Ar, where the concentration of bromine is very low and the concentration of argon in very high. Br₂ is much more likely to collide with an argon atom than with another Br₂ molecule, simply because there is a much higher concentration of Ar than of Br₂.

On the other hand, what Jayron and Keit say about changing the bath gas is correct. If you swamp your Br₂ in N₂ rather than Ar, it changes the efficiency of energy exchange in a collision. See doi:10.1063/1.463803, for example.

If there are two or more "bath gases" present (e.g. mostly Ar with some N₂), one may be more efficient at exchanging energy in a collision. So the most likely collision partner (e.g. Ar) may not be responsible for most of the activation or deactivation. Less concentrated but more efficient quenchers (e.g. N₂) may account for more of the total quenching rate.

My point is this: even for a particular, defined mixture of gases, it is usually impossible to say "M = one particular molecule".

So there you have it. --Ben (talk) 12:22, 25 April 2012 (UTC)

Thanks, Ben, for all your care in posting, however - If it is usually impossible to say what molecule M is, even for a particular defined mixture, then what is the point of reaction rate calculations or Arrhenius data? Yes, the container walls can act as a collision partner, but in practical container sizes, won't the collisions between gas molecules far out-number collisions with the walls? So can't the walls usually be neglected? Presumably, to calculate the overall reaction rate, one needs to evaluate the rate equation for each M possible, including the wall material if necessary? So one must identify each possible M value? Ratbone120.145.51.113 (talk) 13:52, 25 April 2012 (UTC)

With low pressure gas mixtures, container walls might become important.

In practice, M often cancels out when rate equations are combined. It can also be factorised out into a pre-exponential term, which can be ignored for many purposes. --Ben (talk) 14:21, 25 April 2012 (UTC)

Hi, I'm an atmospheric chemist and, as such, use M quite often! Ben's done a pretty good job above explaining so I won't duplicate. One thing, in atmospheric chemistry contexts, the bulk of the atmosphere (the non-trace gas part) is generally invariant enough that it's a reasonable approximation to consider M as any gas (thus [M], the concentration of M, equals the concentration of air) and use a single rate constant derived empirically in the atmosphere. LukeSurl ^{t c} 00:20, 28 April 2012 (UTC)

Yes, that makes sense. Since you are only dealing with trace reactants producing trace products, the concentration of any sort of M won't change significantly. But what rate coeficients do you use? In theory, I guess you could use rate coeficients adjusted for the standard composition of air (about 80% nitrogen, about 20% oxygen) - your empirical value. But given that rate coeficients are generally not published for bath gas = air, and the accuracy of published rate coeficients is pretty rough anyway, couldn't you just take M = N₂? Unfortunately, I am interested in reactions in closed containers where the reaction products approach 100% of what's in the container and all reactants are consumed. So presumably I must evaluate, throughout the reaction duration, each reaction for each value of M, M being in turn each of the reactants and products. Ratbone124.178.39.157 (talk) 02:46, 28 April 2012 (UTC)

Diamminehydridoboron(0) [BH(NH₃)₂]

Is it possible to generate this compound, or will it fly appart like an overfilled closet? Or perhaps rearrange itself and spit out anything that doesn't quite fit, like a box full of mattress springs? Plasmic Physics (talk) 04:31, 24 April 2012 (UTC)

Apparently, diamminedihydridoboron(1+) [BH₂(NH₃)₂]⁺ does exist. Plasmic Physics (talk) 04:34, 24 April 2012 (UTC)

I'm seeing several references to diaminoboranes here, some over 30 years old: [1]. Assuming you mean HB(NH₂)₂. If, instead, you really mean ammineboron complexes, then this search shows similar complexes, such as dihydridodiammineboron(III) ions. That may give you some leads. --Jayron32 04:46, 24 April 2012 (UTC)

Note: Plasmic Physics made additions to his initial posts which make my answers seem irrelevent, but which were relevent at the time given some possible unclear ideas. --Jayron32 16:40, 24 April 2012 (UTC)

Namely, I added formulae in the title and in my second post. Plasmic Physics (talk) 21:50, 24 April 2012 (UTC)

Oxidocarbon(0) [CO] vs. trifluoridoboron(0) [BF₃]

Why on Wikipedia, are carbon monoxide and boron trifluoride both considered to have a bond order of three, even though a fourth bond is delocalizsed over the boron trifluoride molecule? Plasmic Physics (talk) 03:07, 25 April 2012 (UTC)

BF₃ has a practical bond order for each B-F bond of slightly higher than 1. The nominal bond order should be 1 1/3 if following the "octet rule" for boron. this is very approximate, however, as the boron will be highly electron deficient, so the actual electron density within the bonds themselves will be somewhat closer to true single bonds than predicted by lewis theory. Carbon monoxide has a nominal bond order of 3, the expected electron density will of course be slightly less (especially as compared to the isoelectronic N₂ molecule) because of the electronegativity difference between carbon and oxygen, meaning that some of the electron density will be localized to the oxygen rather than "in the bond". As an aside, I'm not sure why you are using metal coordination complex nomenclature for simple covalent molecules, I suppose it works, but it is awkward, and you'll not find a lot of texts using the name oxidocarbon (0) for carbon monoxide. They just use "carbon monoxide", and you probably should too. --Jayron32 03:30, 25 April 2012 (UTC)

Are you saying that the difference is simply an effect of rounding the individual bond orders between atoms within the respective molecules? Wouldn't that mean that the nitrate ion should be depicted as having all single bonds as well? With respect to your question: as you know, I am a black and white, or an eiher/or person, I like to sort everything into nice little boxes. I've decided to use the coordination complex formula notation for any molecule containing that is electron deficient, unless it is itself a ligand. As much as I'd like to, I won't introduce my systems into article pages. Plasmic Physics (talk) 04:13, 25 April 2012 (UTC)

It is black and white: If it isn't a metal-ligand coordination complex, don't use that naming system. Every serious chemist I know, even those that work in the field of metal-ligand chemistry, would consider such use of that naming system for simple molecules like carbon monoxide to be idiosyncratic to the point of being annoying. There is no gray, the naming system is used for metal-ligand complexes alone. Regarding BF₃, you'll find cases in the literature for both extremes. Since the boron in such compounds are almost universally electron deficient, boron is often depicted as having an "incomplete octet", and the lewis diagram for boranes often shows boron to form only three (rather than the expected 4) bonds.See this for one example. Some people are uncomfortable leaving boron with an incomplete octet, when there are fluorines there with lone pairs they could donate and create an octet for boron without losing one themselves; such a structure would have three resonance forms, and a nominal bond order of 1 1/3. this page, for example, takes that tack. Thus, there isn't complete consensus on how to treat BF₃ from a valence bond theory treatment. The physical data for BF₃ is somewhat inconclusive: the gas-phase bond length is somewhat shorter than a true B-F single bond would be expected to be, but it is also somewhat longer than what a 1 1/3 bond would be. If you want the nitty-gritty, I did find this journal article from 1969: [2], which would probably be fairly early work on the electronic structure of BF₃. You could certainly use it to guide your research. --Jayron32 04:50, 25 April 2012 (UTC)

See Talk:Boron trifluoride#Covalent or ionic?. BF₃ is quite ionic, so we may be stretching the truth a little too far by thinking in terms of covalent MOs. Nevertheless: are you comparing BF₃ and NO₃⁻? Think of this diagram. You want to know why the N–O bond order is higher than the B–F bond order. One reason is this: oxygen is less electronegative than fluorine, so it holds onto its electrons less tightly. Therefore oxygen donates pi electrons more easily to an electron deficient centre (to N²⁺) than fluorine does (to B). --Ben (talk) 12:39, 25 April 2012 (UTC)

I'd say it would be more accurate to say that BF₃ is highly polarized; to say it is "ionic" would imply that it actually, you know, forms ions readily, that is there should be B³⁺ and F^1- ions, either in solid form (as an ionic lattice) or in solution. I don't see any evidence that either of these things happens: it is a room temperature gas and doesn't form strongly electrolytic solutions in water. Ionic is as ionic does, and this is not an ionic compound. Heuristic tools used to determine ionic character (such as electronegativity difference or charge distribution) are secondary to actual physical and chemical properties. Instead, it is better to think of it a highly polarized molecule, rather than as distinct ions. The prevailing literature, both journals and student texts, treats it as a molecular substance, so I don't see why we shouldn't do so in this discussion here. --Jayron32 13:37, 25 April 2012 (UTC)

(edit conflict)The difference between ionic and covalent MOs is decided upon a completely arbitrary limit on the wavefunction coefficients, to create non-bonding orbitals. There really is no such thing as non-bonding orbitals, just arbitrarily weak overlap. It's a cse of how many grains of sand makes a pile. I know the true bond orders, I wanted to know why Wikipedia decides to not reflect the correct bond orders in all cases, such as the first two exaples given. Carbon monoxide has a bond order of three and it is reflected by a triple bond between carbon and oxygen, however, boron trifluoride has a bond order of four, yet only three sinngle bonds are indicated. Instead of the fourth being equally divided between the other three B-F bonds, it is completely dropped. Yet, for the nitrate ion, the bonds are depicted as aromatised and not singular bonded. I'm saying that ideally, the B-F bonds sould also be aromatised. Plasmic Physics (talk) 13:52, 25 April 2012 (UTC)

Jayron: I probably went to far, but the ionic argument goes as follows: Inorg. Chem., 1997, 36 (14), 3022–3030 "gives calculated partial charges (B: +2.43, F: −0.81) which indicate BF₃ is predominantly an ionic molecule. The fact that BF₃ does not condense to an ionic solid is explained by "size limitations on the maximum coordination number" that boron can achieve. If BF₃ adopted the AlF₃ structure (ionic solid), boron would be octahedrally coordinated by six fluorides."

See Boron trifluoride#Hydrolysis: does dissolve BF₃ in water, but it undergoes reactions. This is not a simple test of ionicity.

Plasmic Physics: let's be clear, delocalised is not the same as aromatic. And bond order refers to the strength of interaction between two atoms only. BF₃ as a whole molecule does not have a bond order. Each B−F bond has a bond order, often said to be greater than 1.

I'm saying BF₃ is drawn localised whereas NO₃⁻ is drawn delocalised because the latter is more delocalised. --Ben (talk) 14:38, 25 April 2012 (UTC)

The bond order of a molecule is calculated by subtracting the sum of antibonding electrons from the sum of bonding electrons. Plasmic Physics (talk) 23:01, 25 April 2012 (UTC)

I've never seen bond order used that way. Half the difference in the number of antibonding and bonding electrons is the usual way to calculate the bond order of a localised bond. Do you have a reference for your "molecular bond order" definition? --Ben (talk) 11:28, 26 April 2012 (UTC)

That is the way that I was taught in my paper, we use MELDs for this calculation - where it is difficult to isolate MOs of localised bonds. We calculate the bond order of the molecule, and then divide it up by combining MO theory with valence bond theory. Plasmic Physics (talk) 11:39, 26 April 2012 (UTC)

That's more like it. You normally assume the sigma framework follows the Lewis structure, so three single B–F bonds in BF₃. Then you work out any pi bonding with the MO calculations. So your B–F bond order is 1+x, where x is the fraction of a full B=F pi bond calculated to be present. What did you find for x? --Ben (talk) 12:01, 26 April 2012 (UTC)

Average x must be equivalent to ⅓ to make a total bond order of 4. Plasmic Physics (talk) 12:32, 26 April 2012 (UTC)

Simplest being to show affection

Which simplest (taxonomically lowest) living organism is capable to show affection and/or love?--176.241.247.17 (talk) 09:46, 24 April 2012 (UTC)

With all due respect, your question is very problematic. Affection and love are not concepts amenable to biological discussion. We can talk about animal behaviors, but we have very little way of knowing their internal states (which is what love and affection are in humans). You might be interested in Beetle#Parental_care, or Frog#Parental_care, but understand that we cannot say a beetle loves its children because it takes care of them. (sadly parental care is completely human-centric, we need a parental care_(biology)...) SemanticMantis (talk) 12:49, 24 April 2012 (UTC)

You may also be interested in Pair bonding and courtship displays. Both of these are interesting features of animal behavior that we sometimes interpret as being similar to human affection, even though that interpretation has no scientific basis. For a fun example of courtship in "lower" animals, enjoy this clip of a dancing spider [3] :) SemanticMantis (talk) 13:36, 24 April 2012 (UTC)

Perhaps more concisely, affection and love have conotations of conscious thought. Consider certain species of crocodile, where the parent returns to the nest, carefully digs up the hatchings, and carries them to the water. Does the parent crocodile fell affection for her little ones? Who knows? It would seem to be a pre-programmed instinct, and probably it's nothing more that the parent feeling more at ease having done it, maybe not even that. Keit121.215.68.210 (talk) 13:43, 24 April 2012 (UTC)

I don't think most dog owners would claim that their pets are not affectionate, and further complicating matters, individuals vary widely in their temperament. Some people are very selective with their affections, while others are more gregarious. In addition, there are instinctual and environmental processes that underlie our preferences. For instance, identical twins raised apart have often developed eerily similar interests. More generally, within the animal kingdom, the sharing of the many autonomic senses and experiences is as natural as sharing molecules. Thus, even though individuals do not share in everything, animals do tend to share instincts, and its completely futile to ignore the fact that we do have them too. --Modocc (talk) 18:12, 24 April 2012 (UTC)

Social grooming in animals is likely accompanied by emotional states. Bus stop (talk) 18:24, 24 April 2012 (UTC)

Ah, yes, nice link and pics too and this may explain, in part, why I've preferred using a flea comb on my pets to other flea control methods. My pets have liked the extra attention I'm giving them too. :-) --Modocc (talk) 18:37, 24 April 2012 (UTC)

You raise a very good point. Indeed, I do believe that my dog shows affection to me. However, I also know that this belief is not rooted in science. Instead, it is based upon my feelings, emotions, and anthropomorphic interpretations. I suppose there is a somewhat scientific angle from which these interactions could be studied, e.g. interspecies communication. SemanticMantis (talk) 18:56, 24 April 2012 (UTC)

I do think my position is scientific, in part, because our own conceptual models for "being affectionate and experiencing affection" are based on multiple data points with respect to our knowledge of, and interactions with, others. We each know for instance that some people are quite good at feigning affection, nevertheless we can use our knowledge of our shared biology and actions to measure the extent of individuals' affections, perhaps scaled from -10 to 10. In addition, science is about creating and maintaining valid models and the model that many animals experience affection too is simple enough to embrace as opposed to a model that we, as humans, are entirely different. --Modocc (talk) 20:53, 24 April 2012 (UTC)

Yes. I work in a high school, and it's quite normal to see some teenage boys acting very affectionately towards some girls, while all the while their primary aim is to get into their pants. But one can never be certain, can one? It may be true love. HiLo48 (talk) 21:53, 24 April 2012 (UTC)

I didn't mean to say the scientific model was that no animal can experience emotions, nor that I espouse this view. I just meant to stress that scientifically identifying and studying these emotions is extremely difficult (even in humans or chimps, let alone birds, lizards, or bugs). So I tried to provide useful links to similar -and more scientific- concepts. I should have said above "...even though that interpretation may not have a firm scientific basis." SemanticMantis (talk) 23:56, 24 April 2012 (UTC)

The old idea that animals could not experience emotions or true affection was not really rooted in science either. It was certainly influenced by religious ideas about the nature of humans vs. other animals. It's not an easy thing to study scientifically, but Occam's razor certainly points toward some animals having some of the same emotions we do.--Srleffler (talk) 17:56, 26 April 2012 (UTC)

"Taxanomically lowest" (in the original question) is also problematic. By what scale? --ColinFine (talk) 19:32, 24 April 2012 (UTC)

Taking bacteria as being the taxonomically highest because they are at the top of the clade tree (they are also arguably top of the food chain) and everything that has descended from bacteria as being taxonomically lower in heirarchies, then the answer to the original question is human beings. SpinningSpark 11:52, 25 April 2012 (UTC)

• Certainly birds show "affection or love." I have observed a Guinea pig and a canary who were fond of one another (the bird laid eggs for the first time only after the male Guinea pig was housed in a cage near her cage, and they vocalized constantly to one another). When the Guinea pig was removed (returned to school after being housed during holiday), the canary was obviously depressed for a time (off her feed, and less grooming and vocalization). Edison (talk) 04:57, 25 April 2012 (UTC)

When I step on my dog's paw he yelps like I would if my hand were similarly stepped on. When I forget to let him in and it's raining he whines like I would. When I scretch his ear he moans in a way that I would if I were getting a full body massage. When I cried after a serious injury he would come to me as if to say "what can I do to help?". That shows some type of concern, which one might consider care. I colud go on..165.212.189.187 (talk) 13:20, 25 April 2012 (UTC)

In other words, your dog displays behaviors that, if he were human, you would interpret as human affection. That just means you are projecting your own anthropomorphized image of his behavior, and doesn't mean he experiences the internal emotion we humans call affection. Until dogs can tell us what they feel, we're just left with behaviors. There is not necessarily any evidence that a dogs behavior has (or has not)any connection to an internal emotional state in a way that is analogous to humans. --Jayron32 13:15, 26 April 2012 (UTC)

So just explain the whining if its barely audible and the moaning if it is not because it feels sooo good?165.212.189.187 (talk) 16:12, 26 April 2012 (UTC)

Identifying affection from behaviors in nonhuman animals does not just simply imply anthropomorphism, which is a form of overgeneralization, though. Such an error can happen should there be no emotion occurring with a behavior, and people mistakenly believe there is anyway on account of the behavior, and we are rightly skeptical about supposing ants having affections, even if its just a tiny +-.01 bit. Of course, many actors that seem affectionate are not, therefore if we are to believe only what we are verbally told to be true (and ignoring other forms of communication), then one can argue that there isn't any reliable evidence that mans behavior has any connection to are own personal states either. Its a very poor position.. to be unplugged from if necessary when people go further denying that reality exists too because they cannot trust their own senses. Nevertheless, we have much more concrete evidence, from known common cell types to common brain structures, along with various complex behaviors that support the claim that dogs experience affection. --Modocc (talk) 14:40, 26 April 2012 (UTC)

I've seen stories on TV, including but not exclusive to Animal Planet, about various cross-species friendships between animals. One especially touching story had to do with an animal compound or zoo in which an elephant had become fast friends with a dog. The dog was taken to the hospital for surgery, and the elephant waited outside the hospital until the dog was well enough to be let loose again. The joy in both animals was evident.[4] And later, when the dog was killed, apparently by a pack of coyotes, the elephant carried the dead dog back to the main area, and moped for days. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:48, 25 April 2012 (UTC)

Hermaphroditism

Does this NSFW image depict a true hermaphrodite, or just a person with both genetalia? Crisco 1492 (talk) 10:20, 24 April 2012 (UTC)

Well, the museum page shows the person was a hermaphrodite. --SupernovaExplosion ^Talk 11:55, 24 April 2012 (UTC)

According to this paper (p.3), "As early as 1860 the French photographer ... Nadar, took a series of nine photographs – which I take as the first examples of medical pictures of intersex patients ... that depicted a young intersex patient. They show the genitals of a hermaphrodite...". --SupernovaExplosion ^Talk 12:00, 24 April 2012 (UTC)

There is a journal article titled Early Photo-Illustration of a Hermaphrodite by the French Photographer and Artist Nadar in 1860. Hope this clears all the doubt. --SupernovaExplosion ^Talk 12:01, 24 April 2012 (UTC)

• In that case these photographs could have their own article... Imagine the DYK... But if I'm not mistaken, intersex and hermaphrodite are different. Crisco 1492 (talk) 12:51, 24 April 2012 (UTC)

In humans, hermaphroditism is a very specific (and probably non-existent) subset of of the spectrum of intersex conditions. --Carnildo (talk) 02:21, 25 April 2012 (UTC)

The picture should be moved to a file name that spells "genitalia" correctly. - Nunh-huh 02:52, 25 April 2012 (UTC)

True hermaphroditism in humans is not possible. The measure is functioning reproductive systems of both sexes. In humans, the competing hormonal levels would wreck the individual's ability to thrive. What you actually get is someone whose external sexual characteristics resemble both genders, but functionally only one of those sexes works. And I use "functionally" in the loosest terms. — The Hand That Feeds You:^Bite 15:05, 25 April 2012 (UTC)

Our True hermaphroditism article suggests true hermaphroditism in humans normally refers to an individual who has both ovarian and testicular tissue. It also notes that no case is known where both gonodal tissues were functioning but there have been cases where one of the tissues has been, and the individual was fertile. Although as our hermaphrodite article notes with a link to [5], the term is no longer preferred among many in the intersex community and elsewhere and there are more descriptives alternatives like those mentioned in disorders of sex development. In any case, I expect it is fairly unlikely that we have sufficient info on the person in the photo to know if they will fit in to the true hermaphrodite description. (For example Pseudohermaphroditism wasn't cointed until 1876.) Nil Einne (talk) 16:54, 25 April 2012 (UTC)

To become supercentenarian

What measures should be taken by someone in their mid-20s to become a supercentenarian? --SupernovaExplosion ^Talk 13:47, 24 April 2012 (UTC)

Choose your grandparents well. --TammyMoet (talk) 13:55, 24 April 2012 (UTC)

What I've heard is that a good food, good air, a peaceful state of mind, a strong relationship, and a daily mission to do, are all condusive to a long life. Conversely, a poor diet, polluted air, stress, lonesomeness, and boredom/lazyness are condusive to an early grave. Plasmic Physics (talk) 14:01, 24 April 2012 (UTC)

Is that medical advice?165.212.189.187 (talk) 15:37, 24 April 2012 (UTC)

No. See Wikipedia:Reference_desk/Guidelines/Medical_advice#What_does_this_guideline_apply_to.3F. --SupernovaExplosion ^Talk 15:44, 24 April 2012 (UTC)

Be lucky. There are certainly lifestyle choices that will affect how long you are likely to live, but if you're talking about living to 110 or over, then luck is really the biggest factor. --Tango (talk) 18:48, 24 April 2012 (UTC)

A near starvation diet seems to prolong life, but I'm not sure if it's worth it. StuRat (talk) 18:51, 24 April 2012 (UTC)

It is literally a case of "use it or lose it" - people will lose their mind and their strength, the less they use it. The older you get the easier it becomes (to lose), and the longer it takes to return to normal, it may never return to normal. Plasmic Physics (talk) 21:58, 24 April 2012 (UTC)

But you can also overuse your body, resulting in damage. There's an ideal amount of exercise, and either more or less is harmful. Muscle-bound body-builders aren't particularly healthy, for example. This might apply to the overuse of the mind, too, if you're thinking so hard each day you give yourself headaches, for example. StuRat (talk) 22:04, 24 April 2012 (UTC)

I recall an article in Scientific American a few years ago. The authors interviewed various 100+ year old perons in the USA to find out what those persons thought enable their long life. It turned out there were NO common exercise, food, or other common factor, apart from one: Most of them had close relationships with great- and great-great grandchildren, and/or were actively involved in some community project. Well educated people and people working in teaching or research roles in academia tend to live longer. So, what can someone in their mid-20's do? Plan on having many children, get an advanced degree from a good university. Wickwack120.145.51.113 (talk) 00:09, 25 April 2012 (UTC)

I bet none of those people were morbidly obese. StuRat (talk) 01:23, 25 April 2012 (UTC)

Indeed "not doing what they didn't do because that's probably what killed the others" might be a better mentality here than copying. --145.94.77.43 (talk) 13:14, 26 April 2012 (UTC)

Many Japanese people live to old age, while Japanse people who emigrated to the US don't tend to live very long. In a BBC documentary they showed people in Okinawa; what looked to be 70 year old pensioners cleaining their yard, were in reality 90 to 100 year old people. While it is difficult to point to what specifically you must eat, it is easier to point to things that are unhealthy, like e.g. getting your lunch here.

I think the best thing to do is to use supplements as adviced here, except that the reccomendation for vitamin D is too low. The optimal dose for vitamin D is 10,000 IU/day, as that is what you would naturally get day after day, if you were to live like our Stone Age ancestors.

Also, exercise is important, you should make sure your power to weight ratio is at least 4 Watt/kg. This means that if you weigh 60 kg, you should be able to do 240 Watt on a hometrainer, for at least 30 minutes. Note that top athletes typically have a power to weight ratio of 6 Watt/kg or more.

If you are already past the age of 30 and you have not maintained your fitness (you are sort of a couch-potato), then achieving this fitness goal may be too difficult. In that case, you should first build up your fitness level as best as you can and then use drugs like EPO and anabolic steroids for a short time under medical supervision, to boost your fitness level and then stop using these drugs. While your fitness level will decline a bit after you stop, it will remain permanently above what you could have achieved without using these drugs. Count Iblis (talk) 00:54, 25 April 2012 (UTC)

Yowzer! Terrible advice. The Linus Pauling Institute as a source? Really?? Have you seen our article on Linus Pauling and Megavitamin therapy? Vespine (talk) 23:10, 25 April 2012 (UTC)

The institute is named the "Linus Pauling institute", it employs researcher who publish their work in the regular scientific journals, it's certainly not pseudoscientific garbage what they are doing there. Apart from the rather high recommendation for the vitamin E supplementation, there is nothing "mega" about the doses they recommend. Count Iblis (talk) 19:35, 26 April 2012 (UTC)

Most of the world's supercentenarians will have never been near a hometrainer, let alone studied their power to weight ratio. As for using drugs, not a chance. HiLo48 (talk) 20:06, 25 April 2012 (UTC)

That's because they were already over 50 when the first hometrainers were sold in shops. Low testosterone levels are implicated in a whole host of diseases. Keeping your testosterone levels at the level normal for young people as you get older and older, will allow you to exercise and stay fit as you age. Count Iblis (talk) 19:46, 26 April 2012 (UTC)

That sounds more like marketing guff than evidence based. No-one who has followed your advice is yet a supercentenarian, and it's only your faith that says they one day will be. What's wrong with the old approaches that worked? HiLo48 (talk) 20:35, 26 April 2012 (UTC)

SupernovaExplosion, you may find interesting information at http://www.longevityexperts.com/ and http://www.supercentenarian.com/ and http://www.supercentenarian-research-foundation.org/.

—Wavelength (talk) 01:05, 25 April 2012 (UTC)

Thanks everyone for the replies. --SupernovaExplosion ^Talk 03:35, 25 April 2012 (UTC)

There's more to life than just living. Broba (talk) 22:00, 25 April 2012 (UTC)

Photoelectric effect- why Kth shell??

"in Photoelectric effect the electron is ejected from Kth Shell"...the narrow syllebused book dosent say why?..so i m asking why Kth shell why not outermost shell...please explain theoretically and not mathematically...i did read about it sumwhat but it just got over my head..Thankks--Myownid420 (talk) 14:25, 24 April 2012 (UTC)

I'm not sure where that quote came from. It's most likely that an electron will be knocked out of the outer shell, because those are bound more loosely. The "shells" aren't solid like an eggshell, though, they're probability densities. High-energy particles can pass through the outer shells and knock out inner-shell electrons as well. 130.76.64.116 (talk) 20:08, 24 April 2012 (UTC)

The use of kth shell instead of K shell would lead me to believe that your book does not mean the innermost (K) shell, but rather is using the symbol k as some kind of index. You will need to give a fuller quote from the book (or maybe it can be found online from the isbn). If the book is discussing hydrogen or helium, the K shell will, of course, also be the outermost (and only) shell. SpinningSpark 22:23, 24 April 2012 (UTC)

GPS or Glonass which came first

Hi,

Of GPS and Glonass which came first? The corresponding articles do not seem to give a clear picture, of which of the systems had started construction first. — Preceding unsigned comment added by Gulielmus estavius (talk • contribs) 18:30, 24 April 2012 (UTC)

GPS was initially developed in 1973 and became fully operational in 1994, as stated in the GPS article. Glonass was initially developed in 1976 and became fully operational in 1995, according to that article. I'm not sure what confusion you are having; I got that information from the lede of each article. --Jayron32 19:02, 24 April 2012 (UTC)

Per the articles, GPS was developed in 1973, the first test launch was in 1978, the first modern GPS satellite was launched in 1989, and full operational capability (of 24 satellites) was attained in 1994. GLONASS development, meanwhile, began in 1976, with an initial launch in 1982, and a full constellation of 24 satellites reached in 1995. So, GPS was developed on paper first, launched first, and and was fully operational first, by approximately the same margin throughout the process. Note also that GPS attained first major military use during the first Gulf War, in 1990-1991, prior to officially reaching its full capability. — Lomn 19:03, 24 April 2012 (UTC)

Electric motor and heat

I have a simple electric AC fan that seems to take some time to speed up to maximum speed. Normally I would think this was momentum, but after it's been running, if you turn it off, let it stop, and then turn it back up, it quickly returns to top speed. Much quicker than int he first place. I suppose I'm measuring speed based on loudness, but I think that's a reasonable assumption.

My question is why? I suppose the motor operates better at a certain temperature, but I can't find an explanation why in our electric motor article. Any ideas? Shadowjams (talk) 20:43, 24 April 2012 (UTC)

The viscosity of the lubricating oil will change with temperature. The thicker oil, when cold, will make it move slowly. This is most noticeable when you have a tiny motor barely up to the task. Unfortunately, our cold start and warm start articles seem to be exclusively about computers, and hot start is exclusively about aviation. StuRat (talk) 21:35, 24 April 2012 (UTC)

Has the fan been stored for a while? Is it an old fan? It is common for such fans to display this effect, as dust and absorbed impurities from the air increase the viscosity of the oil, also the oil tends to dry out. Also, as the bearings warm up, they expand slightly, making them looser. To fix, put one or 2 drops only of sewing machine oil or light household oil into each bearing. BE SURE to not use too much oil. Keit121.215.39.8 (talk) 23:57, 24 April 2012 (UTC)

trial

Is the Anders Behring Breivik trial on Court TV or any other TV channels or Internet streaming sites. I saw some video clips of the trial on the BBC website and you could see reporters were recording the trial with video cameras. So shouldn't it be live streamed somewhere? — Preceding unsigned comment added by 64.38.226.88 (talk) 22:16, 24 April 2012 (UTC)

Why should it be? It might be, but it might equally be restricted, either for reasons of public policy or for commercial reasons. (This question might have been better asked at the Entertainment reference desk). --ColinFine (talk) 23:20, 24 April 2012 (UTC)

In the UK the TV news channels and programmes are not giving this man much coverage and I saw TV coverage that showed that newspapers in Norway acted together to keep him and reports of his trial from the front pages and inhibit any publicity he may be seeking. I doubt you will find it live streamed anywhere. Richard Avery (talk) 07:09, 25 April 2012 (UTC)

The BBC reported that "Parts of the trial will be shown on television, but the court will not allow Breivik's testimony or that of his witnesses to be broadcast" [6] AlmostReadytoFly (talk) 12:59, 25 April 2012 (UTC)

Hard to figure why this question is in this particular desk. But the obvious reason not to live-cast the testimony of that character and whatever witnesses he might dredge up, is to try and deny or minimize his potential opportunity to turn the trial into a personal forum. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:40, 25 April 2012 (UTC)

The reason given by the court is twofold (I've only found a source in Swedish): one reason is that a live transmission would influence Breivik's testimony, the other is consideration for the victims and their families.Sjö (talk) 17:44, 25 April 2012 (UTC)

Some official information from the Norwegian court authortities here (in English). Jørgen (talk) 20:50, 25 April 2012 (UTC)

Global warming and Mexico

How is global warming likely to affect mexico? Will more people move to high altitudes like Mexico City? Will it be a worse catastrophe for Mexico than it will be for the United States? Will Mexico be able to sue industrialized nations in the World Court?198.189.194.129 (talk) 23:30, 24 April 2012 (UTC)

More polar nations will tend to benefit, while more equatorial and coastal nations will tend to suffer, but weather patterns may change in unpredictable ways, so it's not a certainty for any particular nation. I would guess there would be more hurricanes of greater severity hitting Mexico, and that low-lying coastal areas will suffer frequent flooding, as sea levels rise. The Yucatán Peninsula appears to be the most vulnerable to hurricanes.

The chances of such a lawsuit working are close to zero. StuRat (talk) 23:43, 24 April 2012 (UTC)

Although it could be entertaining to watch Mexico try to sue India and China. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:41, 25 April 2012 (UTC)

Why would that be entertaining? Do you mean interesting, fascinating?---Rich Peterson216.86.177.36 (talk) 00:51, 27 April 2012 (UTC)

Mexico isn't really a low lying country so the rising ocean won't affect Mexico nearly as much as it would affect places like Bangladesh or the Netherlands, Venice or New Orleans. Precipitation is a major worry for Mexico; all areas will see less rain [7]. Not only will there be 5-10% less rain but rain fall will vary more (same concept as flipping a coin, the more times you do it the closer you get to a 50:50 ratio, less flips(rain events) you can get further from the real average. [8] - Also looks like their glaciers are melting - one less possible tourist attraction/source of water for industry/agriculture/people. Heat waves will also have the potential to be hotter, meaning more deadlier for humans, livestock, and even plants will suffer if not even die.

Mexico, like all global nations may also suffer from other nations suffering. Food prices/supply for example is set on a global scale, if agriculture in other parts of the world suffers, whether from increased insect levels, increased plant respiration, increased evaporation of soil water, more severe violent weather events, drought, heat, etc [9], food supply will fall, yet demand will not fall so prices will rise for all humans on Earth. Other nations suffering also means a reduction in possible trading partners; if we did not trade, nations would be forced to move into industries which they are less proficient at than the ones they are currently in.65.95.23.172 (talk) 22:08, 27 April 2012 (UTC)