Wikipedia:Reference desk/Archives/Science/2016 December 17

= December 17 =

Would a man's testicles stop working if his testicles were cut open and all of his seminiferous tubules were surgically removed?
Would a man's testicles stop working if his testicles were cut open and all of his seminiferous tubules were surgically removed?

Any thoughts on this? Futurist110 (talk) 01:29, 17 December 2016 (UTC)


 * Spermatogenesis occurs only in Seminiferous tubules whose surgical removal leaves the man permanently sterile. However Vasectomy is a less invasive alternative. Blooteuth (talk) 02:07, 17 December 2016 (UTC)


 * Would this man's body still produce testosterone afterwards, though? Futurist110 (talk) 03:08, 17 December 2016 (UTC)


 * Possibly. It would depend on how carefully the seminiferous tubules were removed. ( Don't try this at home! ) See Leydig cell.   D b f i r s   08:25, 17 December 2016 (UTC)


 * OK. However, how easy would it be for a doctor/surgeon to successfully do this? Futurist110 (talk) 18:07, 17 December 2016 (UTC)


 * I don't know. I'd be surprised if any surgeon has tried this.   D b f i r s   18:15, 17 December 2016 (UTC)


 * If you DO try this at home, you may become eligible for a Darwin Award. StuRat (talk) 18:38, 17 December 2016 (UTC)


 * Upload pictures or live video if possible...lol. 103.230.104.28 (talk) 20:09, 17 December 2016 (UTC)


 * Given that rendering oneself infertile is the only nonfatal way I can think of to remove oneself from the gene pool, I suspect that the "may" is more of a "will", considering the likely result of trying this and failing! Double sharp (talk) 11:55, 20 December 2016 (UTC)


 * Maybe you could render yourself so hideously disfigured that nobody will come near you ? (Think Phantom of the Opera.) StuRat (talk) 16:53, 21 December 2016 (UTC)


 * You are considering the surgical removal of 600 metres of tubule, with an average diameter of less than one tenth of a millimetre: that would be well beyond the capacity of even the most able micro-surgeon. Wymspen (talk) 16:30, 21 December 2016 (UTC)


 * Testosterone is produced by Leydig cells, which you might call part of the "sauce" for seminiferous tubule spaghetti. Leydig cells will work for testosterone production if transplanted into castrated male mice.   So it seems likely that some kind of magic seminiferous tubule remover could leave testosterone secretion intact; whether this would have harmful side effects or encounter "interesting new biology" of course remains an open question. Wnt (talk) 12:36, 18 December 2016 (UTC)

Waterfalls get narrower at the bottom?
I observed an approximately 1-m-high waterfall today and wanted to estimate its flow rate. I realized that I could obtain a rough estimation by estimating the cross-section of the waterfall and the water velocity (estimating the latter using the difference in potential energy between the bottom and top of the cascade), and then (assuming my velocity is the average over the whole cross-section), multiplying these quantities. But I realized that, assuming the flow is laminar (it looked as such, but this is quite a strong assumption as it would later turn out!), the cross-sectional area of the water stream would have to decrease as the water descended, by a (naive) application of the continuity equation, because the volume of water flowing through any given cross-section would have to always be the same (neglecting evaporation) as any other cross-section of the waterfall. Larger-scale waterfalls like Vernal Fall hardly shrink in width at all as the water descends. I further conjectured that to maintain laminarity, intermolecular forces must act to pull the water into a smaller stream, so in tall waterfalls, laminar flow cannot be maintained, and the water must fragment as it descends, even if there is negligible wind.

This is my naive interpretation of the fluid dynamics of this situation. How correct is it?--Jasper Deng (talk) 06:10, 17 December 2016 (UTC)
 * Here is a waterfall that spreads as it falls. This one falls pretty much straight over a cliff, but it also spreads before it hits the pool below it. How a waterfall flows is likely to depend on the rock structure around it. ←Baseball Bugs What's up, Doc? carrots→ 09:41, 17 December 2016 (UTC)
 * Here I am assuming a free-falling waterfall, like the one I observed.--Jasper Deng (talk) 09:46, 17 December 2016 (UTC)
 * I'm not sure you're going to find very many truly free-falling waterfalls. Even if they have a "lip" to go over initially, that lip will tend to erode over time, often cutting most deeply into the middle of the stream, thus concentrating the stream and eroding even further in the middle. That's what appears to be happening with Punch Bowl. ←Baseball Bugs What's up, Doc? carrots→ 09:56, 17 December 2016 (UTC)
 * My question is about fluid dynamics under the assumption of laminar free-falling water flow, not about the shape of most waterfalls. The assumption does not apply in the examples you cited; like I said, it most likely can't apply to waterfalls beyond a certain height.--Jasper Deng (talk) 10:32, 17 December 2016 (UTC)
 * I'm not sure why you don't just go to the kitchen tap. It produces a nice little curve just as you described; assuming (dicily) cylindrical symmetry you might take a picture of it, measure pixels, and compare to your math.  As long as it doesn't have one of those mesh screens to insert air to keep it as a cylindrical prism - if it does, just get rid of that and try again. Wnt (talk) 17:53, 17 December 2016 (UTC)


 * Yes, a laminar falling stream must narrow as it accelerates.   Once it narrows too much, Plateau–Rayleigh instability sets in and causes the flow to break up into droplets.  The transition to droplets is why many tall waterfalls appears to turn into white spray before they reach the ground.  Once there is a lot of air mixed into the flow, the terminal velocity is quickly reached and the water stops accelerating.  Dragons flight (talk) 10:33, 17 December 2016 (UTC)
 * It sounds like a waterfall truly exhibiting Laminar flow would have to be pretty small, as with the OP's original example. ←Baseball Bugs What's up, Doc? carrots→ 10:37, 17 December 2016 (UTC)

Chemical ratio
Is there a name for the property chemical potential per mole?


 * Chemical potential is usually measured relative to another chemical potential. In theory, you can separate all the atoms and compare the potential relative to that, but doing so in practice could be difficult.  In general the term is enthalpy.  The Gibbs free energy (which takes into account entropy, a property by which temperature generates disorder) is often of more interest; it explains how endothermic reactions are possible.  An example of a change in enthalpy that is often measured is the heat of hydrogenation (hmmm, see hydrogenation), which does not separate all the atoms but gives a sense of the strength of double bonds.  Another is heat of combustion (measured in a bomb calorimeter, perhaps) that gives a sense relative to known small compounds. Wnt (talk) 17:49, 17 December 2016 (UTC)
 * I'm looking for the chemical property that maps to the inverse of electrical capacitance in a bond graph analysis. That is, the inverse of chemical capacitance.  None of the variables you mention seem to fit that bill.  See this table.  The quantity I want is an effort variable over a displacement variable.  That is chemical potential over integral of molar flow rate according to that table.  This page describes the quantity I want but doesn't name it.  My question is does anybody give this a name. SpinningSpark 18:38, 17 December 2016 (UTC)
 * Oh-- I was totally off here. I was thinking of chemical potential energy from covalent bonds, but you have in mind what I think is the Gibbs free energy corresponding to the entropy that can be extracted by mixing a solution thoroughly.
 * To start with, the inverse of electrical capacitance is (according to farad) J/(V^2) or (C^2)/J. I don't know what to do here with the electrical units ... maybe the eV/e formulation of the volt is most relevant?  It all comes down to energy per ... something, or something per energy.  I may be spinning wheels on your analogy.
 * Going back to chemical potential, I followed the link to Gibbs–Duhem equation - it is interesting that there the infinitesimal increases in chemical potential are multiplied by moles, which implies they are per-mole values. Later on the article says "the chemical potential is just another name for the partial molar Gibbs free energy".  I'm going to go with that as a blind guess, but at this point I should admit I blundered into the wrong lecture hall, duck and run. Wnt (talk) 03:13, 18 December 2016 (UTC)
 * Chemical potential is indeed in units of energy per mole (at least as I understand it, I am an electrical engineer) so the sought quantity has units of joules/mole2. This is exactly analogous to the electrical case.  Capacitance (Q/V) has units of coulombs/volt and inverse capacitance volts/coulomb.  But the volt is equivalent to joules/coulomb (Joule's law) so inverse capacitance has units of joules/coulomb2. SpinningSpark 14:06, 18 December 2016 (UTC)

Biological organism development stages
Article sought. 103.230.105.20 (talk) 19:49, 17 December 2016 (UTC)


 * Ontogeny. μηδείς (talk) 20:24, 17 December 2016 (UTC)
 * Also perhaps Evo Devo for some context on how ontogeny fits in with phylogeny. SemanticMantis (talk) 21:59, 18 December 2016 (UTC)

Feynman Lectures. Lecture 41
It is not clear, the probability of what is $$e ^ {-\text{P.E.}/kT}$$ in classical mechanics? Is it the probability of finding a particle with energy $$\text{P.E.}$$? Then I can't see the difference and the problem. Remaining in classical mechanics we can represent probability as probability of lower energy too ,e.g. $$P(\text{P.E.})=e ^ {-\text{P.E.}/kT}, P(\text{P.E.}+\Delta)=e^{-(\text{P.E.}+\Delta)/kT} \therefore P(\text{P.E.}+\Delta)=P(\text{P.E.})\cdot e^{-\Delta/kT}$$.

And to find total energy we must calculate something like $$\sum p\cdot E$$ (or maybe integrate because in classical mechanics energy is not discrete), where $$p$$ is probability of energy $$E$$.

Username160611000000 (talk) 21:05, 17 December 2016 (UTC)


 * See Boltzmann distribution. I'm not going to say more lest I have it wrong myself... Wnt (talk) 23:23, 17 December 2016 (UTC)


 * Thank you for link. I have read, but it haven't made the subject clearer. On the Lecture 40 summary  photo Feynman wrote that $$e^{-P.E./kT}$$ is the factor to which the probability is  proportional per unit volume. For example we have the probability 50 %, what this means? Is it mean that in 1 cm³ we can find 1 molecule in a half of tests?


 * Or for example we have 1000 atoms in 1 m³ on average (or 10 in 1 cm³) at temperature 273 K. How to calculate probability at zero energy? But if temperature is 273 K then energy must be $$\tfrac{3}{2}NkT = \tfrac{3}{2}1000\cdot 1.38\times10^{-23} \cdot 273$$. Username160611000000 (talk) 09:40, 18 December 2016 (UTC)


 * For more familiar circumstances I'd use this line from the table in the article: "0.0019872041(18) 	kcal/(mol⋅K) - per mole form often used in statistical mechanics - using thermochemical calorie = 4.184 joule"  Note this is the year Budd Dwyer became an hero, in millions.  You take that by some temperature, usually 300 K, and get in that case 0.596 kcal/mol.  Now you know that if an excited state is at +5 kcal/mol, say, it is e^(5/0.596) less likely to happen than the ground state.  In the textbook cases I know of, you're then trying to decide whether that state exists at a high enough frequency to radiate a photon on occasion, or to participate in a chemical reaction.  It's less clear to me how you decide whether, if that's a non-bonded state, the bond actually breaks.  I'm less clear what you're trying to do in your example though, and I can't say this is something I know that well to start with.  Wnt (talk) 12:53, 18 December 2016 (UTC)
 * Ok. Another question. Suppose we have N atoms and they occupy volume of V m³, no field is present. How to calculate the probability of finding (N/V) atoms in 1 m³? I don't think it is N/(V•N) as Feynman writes after Eq. (40.11) . I feel it must be greater, as (N/V) atom/m³ is average density.  The probability to find a certain atom (e.g. atom #41) in 1 m³ is 1/V, because the atom can be in each cubic meter with equal probability. The probability not to find this atom is 1-(1/V). Then the probability not to find all N atoms in this certain m³ is [1-(1/V)]^N. And required prob. = 1-[1-(1/V)]^N. Is it correct? Username160611000000 (talk) 04:49, 20 December 2016 (UTC)


 * Or maybe $$\underbrace{(1-[1-(1/V)]^N)(1-[1-(1/V)]^{N-1})(1-[1-(1/V)]^{N-2})\cdot ...}_{N/V \text{times}}$$ ??
 * Username160611000000 (talk) 05:24, 20 December 2016 (UTC)


 * I don't see your N/(V•N) in the section linked. Dragons flight (talk) 09:16, 20 December 2016 (UTC)
 * Quote:"since P1=n1/N and P0=n0/N." And n = N/V. Username160611000000 (talk) 13:33, 20 December 2016 (UTC)
 * So $$e ^ {-\text{P.E.}/kT}$$ is not the probability. We have some probability according the formula $$\underbrace{(1-[1-(1/V)]^N)(1-[1-(1/V)]^{N-1})(1-[1-(1/V)]^{N-2})\cdot ...}_{N/V \text{times}}$$ and with height increases  we simply multiply by the factor $$e ^ {-\text{height*constant}/kT}$$. Username160611000000 (talk) 15:25, 20 December 2016 (UTC)
 * So $$e ^ {-\text{P.E.}/kT}$$ is not the probability. We have some probability according the formula $$\underbrace{(1-[1-(1/V)]^N)(1-[1-(1/V)]^{N-1})(1-[1-(1/V)]^{N-2})\cdot ...}_{N/V \text{times}}$$ and with height increases  we simply multiply by the factor $$e ^ {-\text{height*constant}/kT}$$. Username160611000000 (talk) 15:25, 20 December 2016 (UTC)