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May 6

What happens if long enough rope is on earth?

49.135.2.215 (talk) 01:26, 6 May 2016 (UTC)Like sushi[reply]

What? Evan ^{(talk|contribs)} 01:27, 6 May 2016 (UTC)[reply]

Earth destroyed by black hole. Sagittarian Milky Way (talk) 02:06, 6 May 2016 (UTC)[reply]

Didn't somebody ask this like a week ago? ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:08, 6 May 2016 (UTC)[reply]

You can use it for a space elevator. It had better be a really strong rope though. --71.110.8.102 (talk) 06:15, 6 May 2016 (UTC)[reply]

I imagine it will be like the infamous "piece of string" - always just slightly too short for whatever it is needed for. 81.132.106.10 (talk) 09:03, 6 May 2016 (UTC)[reply]

If a long rope is on Earth, well, a long rope is on Earth. ThePlatypusofDoom (Talk) 12:29, 7 May 2016 (UTC)[reply]

The Earth will hang itself.--Heron (talk) 07:36, 8 May 2016 (UTC)[reply]

pressure enough to make "carbon string"?

(I can not be watching)

49.135.2.215 (talk) 01:28, 6 May 2016 (UTC)Like sushi[reply]

Read Carbon fibers, Carbon nanotube and Linear acetylenic carbon. High pressure may not be required. Graeme Bartlett (talk) 11:18, 6 May 2016 (UTC)[reply]

Edible-nest swiftlet

Our article on the subject is quite poor. I was wondering if anyone knows the average weight of each nest?Lihaas (talk) 10:17, 6 May 2016 (UTC)[reply]

The German article (de:Weißnestsalangane) says: "Der Durchmesser beträgt ungefähr 6, die Höhe 1,5 Zentimeter, das Gewicht liegt bei ungefähr 14 Gramm." (diameter 6cm, height 1.5 cm, weight around 14 grams). --Wrongfilter (talk) 10:31, 6 May 2016 (UTC)[reply]

Thanks

Man, it takes a heck of a lot to make 1 kg...which sells for some $2000.Lihaas (talk) 13:41, 6 May 2016 (UTC)[reply]

So about 28 cents apiece: have a look at what they have to do to collect them. Alansplodge (talk) 17:38, 6 May 2016 (UTC)[reply]

Isn't that 28 dollars apiece ? StuRat (talk) 03:55, 7 May 2016 (UTC)[reply]

D'oh! Quite right StuRat, $28 each. Still, I suspect that's the wholesale price rather than what the chap at the top of a precarious ladder gets. Alansplodge (talk) 13:35, 8 May 2016 (UTC)[reply]

If they paid better, maybe they could afford sturdier ladders (or how about a cherry picker, or at least a net ?). StuRat (talk) 23:24, 9 May 2016 (UTC) [reply]

I like how the video clearly states that the nest has no nutritional value and adds no flavor. I bet it's a heck of an interesting texture though :) SemanticMantis (talk) 17:49, 6 May 2016 (UTC)[reply]

I dunno about the youtube video but it makes sense at about 30 or so dollars...and its apparently rich in protein but an acquired taste.Lihaas (talk) 12:42, 7 May 2016 (UTC)[reply]

@Alansplodge: Theyre farmed too. (hence easier)Lihaas (talk) 13:06, 8 May 2016 (UTC)[reply]

Thanks Lihaas. Gosh, I don't know how I missed swiftletfarming.blogspot. Alansplodge (talk) 13:35, 8 May 2016 (UTC)[reply]

Extracellular sodium channel blockers vs. intracellular sodium channel blockers

Sodium channel blocker lists both extracellular and intracellular classes of these substances. I noticed that the intracellular forms include several crucial drugs, including a number of WHO-EM drugs (lidocaine, bupivacaine, procainamide, quinidine, phenytoin, carbamazepine), whereas the extracellular forms are some of the most lethal neurotoxins known (saxitoxins, neosaxitoxins, and tetrodotoxins). It would seem that blocking one side of the sodium channel versus the other makes a significant difference, but what exactly is that difference? —/Mendaliv/^2¢/_Δ's/ 20:39, 6 May 2016 (UTC)[reply]

Excellent question! Two thoughts. First, the sodium channel that both TTX and lidocaine act upon is a voltage-gated sodium channel. This means that, unlike the ligand-gated nicotinic sodium / potassium channel, the voltage gated sodium channel is not meant (evolved) to be activated / deactivated / inactivated / deinactivated by a neurotransmitter (ligand) molecule binding to it from the outside. So, whatever molecule binds to it from the outside and blocks it, is either a toxin or a drug, and not a part of a natural physiological processes taking place in the organism. Second, there is no strict way to distinguish between a "poison" and a "drug" - it is purely a question of dosage and intended (or unintended) effect. Toxins are poisons produced by living organisms. It so happens that TTX is used by some fish and amphibian species to protect themselves; so it is ingested by the predator and acts on the predator's neurons from the outside instead of being transported into the neurons first. There are toxins that are transported into the cell, too: for example colchicine or taxanes that interfere with cell division, or Tetanus toxin that blocks neurotransmitter release into the synaptic cleft. But for a fast-acting toxin your best bet is to block the action potential generation and propagation or synaptic transmission from the outside of the neurons, which is what most fast-acting toxins do. --Dr Dima (talk) 22:13, 6 May 2016 (UTC) Continued: as to why the voltage-gated sodium channel "blockers" (actually, mostly modulators) found in medical use bind to the intra-cellular sites of the sodium channel - it may be by chance (novocain was discovered in 1905, long before the sodium channel itself was discovered). Or it may be because binding to the intra-cellular sites allows to modulate, rather than physically block, the channel, which may be advantageous in medical applications. --Dr Dima (talk) 22:50, 6 May 2016 (UTC)[reply]

What I can say for sure so far is that an answer for this one won't be easy. There seem to be some different effects of the two types of drugs, [1] and allegedly (I don't know if this is reproducible) the extracellular toxins also do something to calcium channels ... something subtle. [2] The local anaesthetics affect TTX-resistant channels. [3] There are literally hundreds of papers in PubMed that discuss tetrodotoxin and lidocaine somehow or other, and they're not the easiest papers to read... (Here's another one that might be informative, modeling extracellular access of drugs, but it is not finger food) One trivial thing is TTX is simply longer-acting, [4] but that doesn't really explain it when a drug is taken with the intent of long-term inhibition. Hmmm, hit #249 is a goodun: [5] says "The contrasting patterns of effects between TTX and local anaesthetics suggest that blockade of TTX-sensitive sodium channels alone may not be responsible for the effects of cocaine, lignocaine and benzocaine." That's kind of ancient, but there is TTX-R stuff about lidocaine coming out to the current day: [6] Wnt (talk) 00:18, 7 May 2016 (UTC)[reply]

Very interesting and informative responses, thank you both! So my take on this is that the intracellular versus extracellular distinction might not be the crucial distinction. For instance, it looks like STX and TTX are voltage-insensitive (whereas most pharmaceutical Na+ blockers seem to be voltage-sensitive). Related to this seems to be the affinity, with STX/TTX having very high affinity. Taking off something Dr Dima said above—noting that the distinction between drug and toxin is a fairly arbitrary one—I began to wonder if the problem with STX/TTX is one of therapeutic index... or more simply that useful doses would be extraordinarily small. I saw one article suggesting the use of either STX or TTX for bladder pain syndrome, with results in a human study of anesthesia lasting around 90 (!) days, for example, with a dose measured in micrograms. —/Mendaliv/^2¢/_Δ's/ 03:43, 9 May 2016 (UTC)[reply]

Magnetic quantum number

I recently did some major rewriting at magnetic quantum number, and have a leftover sentence about Larmor precession. The problem is, I have no idea how to relate it to the article. As best as I could figure out looking at a few web sources, the Larmor frequency is a function of the atom, not the m of a particular electron therein, and so electrons with any given m all precess at the same frequency?

Bonus question: I was kind of wondering where the angular momentum goes if multiple photons hit an atom. For example, you have a hydrogen atom with bright multiple frequencies of light shining on it. The first photon carries sqrt(2) reduced Planck units of angular momentum and knocks the atom from its base state (0) to the first level (sqrt(1*2)); all the angular momentum is conserved. But any photons after that knock it up by a smaller amount, not much more than 1, because sqrt(l(l+1)) approximates to l+0.5. Now I suppose this is reversible - it could emit back out photons with the same total angular momentum - and so in a sense the momentum is conserved. But the vector isn't! Or am I missing something... (I'm thinking this is something about how random angular momentum adds up ... but in which direction(s)?)

Anyway, I'd much welcome some eyes at magnetic quantum number to see if I screwed it up badly. Wnt (talk) 21:29, 6 May 2016 (UTC)[reply]

Also, I think there is no way to tell by looking at the shape of the orbital whether it has a +m or -m (but you can tell the absolute value of m). This would be the equivalent of seeing a map of a planet's orbit and not knowing from that alone which way the planet is headed. But can someone confirm with a source?

More generally, it would be nice to get a more intuitive feeling here. If you look at a hydrogen and you see it has a p_x electron, that electron has an angular momentum around the z axis. Does this truly mean that the electron is orbiting in some sense, i.e. that over time it is moving either clockwise or counterclockwise about the z axis? Now there are some real howlers about the meaning of the + and - lobes of orbitals online, but my understanding is that they represent the phases of a wave, and that in intuitive terms this means - correct me if I'm wrong, and this sounds wrong! - that if we somehow could know where the electron really was in the orbital without disturbing its momentum, then in some short time (how short?) it would fall through the nodal plane and turn p in the other sign of lobe in something like half the Bohr orbital period (half of 150 as, according to page 9, but does this get larger with principal quantum number? I'd guess so, and that the extra time corresponds to the extra little lobes near the nucleus in 3p orbitals and up, which it has to pass through?) And so I take the angular momentum to possibly indicate that the electron in a p orbital jumps from lobe to lobe via a course that goes a particular direction around the nucleus? But at some point the analogy seems to break down since the electron jumps over a nodal plane, something with no apparent particle-mechanics analogy. Wnt (talk) 14:29, 7 May 2016 (UTC)[reply]

It "jumps over a nodal plane, something with no apparent particle-mechanics analogy"? Exactly! Because it's a wave-like/particle-like entity, not quite either one. Even on a plucked guitar-string, one can easily see stationary points. If one is very careful (and my quantum-mechanics prof was awesome at this!) one can lightly place a finger at half the guitar's length, pluck the string, and then remove finger...both halves of the string vibrate, and the guitar is stably emitting an octave higher than normal. The hand-wave(ha!)-ing explanation is that the electron (if you are still thinking it's a particle that has discrete positions at specific times) can tunnel across the node. The wave explanation is that there is just a node on the single wavefunction, so "the wave" is on both sides simultaneously, or the string need not "move" the middle in order to be moving on both sides. wikiquote:Quantum mechanics has some good comments from experts about how often QM does not make self-consistent macroscopic/real-world sense. DMacks (talk) 21:03, 7 May 2016 (UTC)[reply]

@DMacks: I recognize that many sources say QM is hard to understand. But I remain unsure when and how much those limits can be pushed. What emboldens me most is that alternating current can be written in similar crazy notation, with phase angles and complex numbers. Yet it has an intuitively comprehensible basis. So I feel like I shouldn't give up trying to find one here. I am thinking (not sure) that if you look at the f orbital diagram I added to the article, that you can sort of see how an f orbital with m=0 is vibrating up and down between lobes in the z-axis, and one with m=1 is orbiting in something like an x-z plane from lobe to lobe, with only a slight motion around the z axis, whereas the m=3 case is more or less the same orbit, but now totally in the xy plane so that all the angular momentum is around the z axis. The motion from lobe to lobe (or node to node) would seem to mark some kind of path...

Which gets me to another question, I suppose. If you "lightly measure" a population of electrons to be within one particular lobe of an f orbital, say, can you measure them within attoseconds later and see that they may move to the adjacent lobes, but not all the way to the lobe on the far side of the atom? Wnt (talk) 01:02, 8 May 2016 (UTC)[reply]

For an analogy of the "is the electron moving or not" question, consider a flashlight/laser beam shining continuously on an absorbent (dark) surface. The light has momentum (pointing along the direction of the beam, toward its destination). But the situation is also time-independent. Is the light moving? In some sense yes, in some sense no. It's the same question for the electron orbits. That it's a quantum wave function, and the motion is circular instead of linear, doesn't matter as far as this question goes. The electron orbits in the same sense that the light moves.

For a classical analogy of "jumping through nodes" consider a standing wave. You can think of a standing wave in one dimension as a superposition of left-moving and right-moving waves. Is the light (or whatever) moving left and right through the nodes? I don't know, but whatever your answer is in the classical case, that's the answer in the quantum case too. -- BenRG (talk) 17:46, 8 May 2016 (UTC)[reply]

Maybe I'm being simple-minded here, but I feel like if an electron has energy and angular momentum it has to be moving. Still, the whole question may be complicated. It seems like one guy at the center of what has been published lately on attosecond transient absorption spectroscopy, who was looking at things like the transition of an electron from 4d orbitals to Rydberg atoms, says "Such a noble-gas ion which inherits a hole in the valence p_z orbital is not anymore in an eigenstate but rather in a superposition of different quantum states." And what he can do seems a lot blunter than what I wanted, looking at where an electron is in an orbital and then looking a few attoseconds later, which for now still seems like nothing but a thought experiment AFAICT. This is all far beyond my understanding. Honestly, I'm still not really understanding what that ring diagram I added above really means - the complex phases are normally shown as spaces between orbital lobes, but in this series of images, even for p orbitals (except along the z axis! due to multiple possible orientations superimposed?), they are shown as linkages, probably the very "orbits" I was looking to see. But dang it, if only I got what the e^itheta in these solutions "really means", and comprehended how these segments are bricked together to build up orbitals, I feel like a lot of quantum mechanics would be very intuitive. Wnt (talk) 20:16, 8 May 2016 (UTC)[reply]

No one knows why nature chose complex-valued wave functions over real-valued or quaternion-valued or something else, so you're in good company if you don't understand that.

It's important to understand that these orbitals are not the only states the electrons can be in, they are just an orthonormal basis (linear algebra) for the states. Any orthonormal linear combination of those states would also work. In commons:Hydrogen orbitals 3D#n = 2, they've chosen a different set of 2p basis vectors than you usually see in these collections of pretty pictures. Usually, they would use the middle one, the sum of the left and right ones, and the difference of the left and right ones (times a normalization/phase constant). The left and right ones are e^iθ and e^−iθ times a function of r and φ, and their sum and difference are cos θ and sin θ times that function of r and φ (still ignoring the constant factors). cos θ and sin θ have two peaks each, of opposite complex phase (+1 of phase 0 and −1 of phase 180°), and those two phases are colored blue and yellow in these pictures. So the sum and difference of the left and right pictures would look like the center picture turned on its side, which is what you usually see in pictures of the 2p orbitals (e.g. here). It's not the same functions drawn in different ways, but different functions (spanning the same space) drawn in more or less the same way.

All of these orbitals taken together are a complete basis for bound states, so any bound electron wave function is a linear combination of them—even one where the electron is localized, and very far from the atom, and basically orbiting it classically, as in a Rydberg atom. But these approximately classical states are combinations of basis orbitals from different rows of commons:Hydrogen orbitals 3D, which means they don't have a definite energy or angular momentum. The guy you quoted misspoke when he said "a superposition of different quantum states". He meant "a superposition of eigenstates of different quantum numbers", i.e., a weighted sum of basis functions from different rows of commons:Hydrogen orbitals 3D. -- BenRG (talk) 05:29, 9 May 2016 (UTC)[reply]

@BenRG: I suppose I still don't fully understand the superposition process and how it affects angular momentum. If I simply look at one of those looped p orbitals, I can imagine that the complex phase simply means when the electron is most likely to be at a particular point, almost like with alternating current. It is around the position with a +1 value at t=0, around the position with +i at t=a/4, around -1 at t=a/2 etc., where I would expect a ought to be some small multiple (1?) of the Bohr period. And if all that were true, then I see it having an angular momentum to match m=1 or m=-1 (per the right hand rule). However, if I sum these two together, getting the yellow and blue spheres, now it's just going back and forth straight along the x or y axis, and it has no angular momentum. Now of course I pretty much have to sum them together if I expect the electron to be located somewhere in particular, like pointing at another atom, but don't know which way it's moving - I think that's telling me the angular momentum is in a superposition of eigenstates in those instances? If that's true, then the yellow-and-blue version for m=0 should be taken to mean that we know the angular momentum around z is zero, and we don't know which way it is in the other axes. But is any of this right? And if it is, might we (by which I'm afraid I mean you, sorry!) get some sources and modify one of the articles to somehow explain the difference between the ring and the yellow-and-blue pictures of an orbital? Wnt (talk) 11:30, 9 May 2016 (UTC)[reply]

Looking through [7], it does appear possible to deduce m by looking at the orbital. The absolute value of m is equal to the number of red bands (phase = i) in any torus for a given orbital (and zero for the blue-and-yellow solutions on the page). The sign of m is negative if blue, purple, red run clockwise in any given torus, and positive if they run counterclockwise. I suppose a planetary orbit would also provide the sign of its angular momentum, if it were drawn as a complex number in which the e^itheta angle is a function of time. Wnt (talk) 15:07, 9 May 2016 (UTC)[reply]

Alright, to bite the bullet here, in a Bohr model, FWIW:

v={\sqrt {Zk_{\mathrm {e} }e^{2} \over m_{\mathrm {e} }r}}.

For now I'll say it's an actual hydrogen (Z=1), leaving me to work 8.9875517873681764×109 N·m2/C2 * (−1.602176565(35)×10−19 C)^2 / 9.10938215(45)×10−31 kg. Uck... comes out, in the unlikely event I did not foul up, to 253.3 N m2/kg = 253.3 m3/s2. Divide by r, take the square root, that's m/s. For the Bohr radius of 0.0529E-9 m, I get the value for v is 2188058 m/s. [D'oh! That's c * alpha! Apparently Coulomb's constant * electron charge^2 = reduced Planck constant * speed of light * fine structure constant.] To go around the Bohr radius then is 24.18 attoseconds ... no, times 2 pi, because it's the circumference ... 151.9 attoseconds. Caveat being that the ground state in the Bohr atom has angular momentum with n=1! Well, I'm starting where I can. Anyway, the classical period is expected to be proportional to n^3, the angular momentum to n, the velocity to 1/n, the radius to n^2 ... so the time to cross each segment between two red bands in high-m orbitals like the graphic above would be inversely proportional to n^2. (Not to m, but for the highest-m orbitals that look like a single torus around the z axis, m is equal to l is equal to n) So clearly, unlike what I was first wondering, the electron isn't simply vibrating at some base Compton rate as it goes around one of these tori. (According to Compton wavelength the classical radius of an electron, FWIW, is 2.8E-15 m, much smaller than any of these features, while the other measure is simply the Bohr radius) The ionization energy for the innermost orbital is -13.6 eV, vs. 510 keV electron mass, so the Compton wavelength of that negative energy, if there were such a thing, would be 37500 times higher, or 1.05E-10 meters ... hmmm, that's close to the 3.323E-10 meters of the Bohr circumference. Especially curious since that should go up proportional to n^2, just like the length of the circumference in the higher-energy orbitals, because the ionization energy drops per n^2. Is there anything in physics about these orbitals somehow reflecting the Compton wavelength of the ionization energy? That's weird! Wnt (talk) 21:31, 9 May 2016 (UTC)[reply]

@Wnt: The orbitals do correspond roughly to certain classical point-particle orbits. The n−1 = ℓ = ±m orbitals are like circular orbits. The ℓ = 0 orbits are like degenerate elliptical orbits with a minor axis of 0 and one ~~focus~~end at the nucleus. Intermediate values of ℓ correspond to intermediate eccentricities. (Analogues of parabolic and hyperbolic orbits also exist, but they aren't included in the standard solution to the hydrogen atom since it considers only bound states.)

The |m| ≠ ℓ orbitals are like sums (superpositions) of |m| = ℓ orbitals with different orientations. In particular the ℓ = 1 (p), m = 0 orbitals are like a sum of opposite circular orbits around any axis perpendicular to the z axis. A simpler analogue of this (in classical or quantum physics) is a standing wave in a one-dimensional wave guide with conductors at each end. It's a sum/superposition of left-moving and right-moving waves, and the particle analogue is a particle bouncing back and forth off of the ends, alternately moving left and right. I think it's a bad idea to think of the particle as having some location the wave (which would require it to cross internal nodes, among other things). Rather, the wave represents a kind of smearing of the particle over its entire periodic path. The wavicle is everywhere at every time, but still moving (half of it left and half of it right), like two flashlight beams pointed at each other. -- BenRG (talk) 16:46, 10 May 2016 (UTC)[reply]

Does petting an human release endorphins?

In the past I've heard explanation why people are relaxed while getting petting, but I forgot the explanation. Is it because that the body release endorphins? (I know for sure that it releases the hormone oxytocin but I'm not sure about the endorphins) 93.126.95.68 (talk) 23:28, 6 May 2016 (UTC)[reply]

Probably more so than with your average dog or cat. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:38, 6 May 2016 (UTC)[reply]

Apparently so. Complementary Psychosocial Interventions in Child and Adolescent Psychiatry: Pet Assisted Therapy--Aspro (talk) 23:45, 6 May 2016 (UTC)[reply]

If I'm nor mistaken, this article deals with petting pets rather than petting human. Am I right? 93.126.95.68 (talk) 23:54, 6 May 2016 (UTC)[reply]

Am I right or mistaken that humans are animals also!--Aspro (talk) 00:32, 7 May 2016 (UTC)[reply]

"petting" is a fairly antiquated term for physical affection...like "heavy petting" instead of "making out"...the OP is using the term in this manner...68.48.241.158 (talk) 01:51, 7 May 2016 (UTC)[reply]

I doubt it releases Oxycontin. You were probably thinking of oxytocin. --71.110.8.102 (talk) 04:46, 7 May 2016 (UTC)[reply]

Thank you, I corrected it. 93.126.95.68 (talk) 12:20, 7 May 2016 (UTC)[reply]

According to Dr Nor Ashikin Mokhtar, there are many hormones and neurotransmitters released during foreplay but oxytocin and endorphins only seem to be released during an orgasm. Richerman (talk) 08:00, 7 May 2016 (UTC)[reply]

According to the article oxytocin "In a 2003 study, both humans and dog oxytocin levels in the blood rose after five to 24 minutes of a petting session."93.126.95.68 (talk) 12:22, 7 May 2016 (UTC)[reply]