Wikipedia:Reference desk/Archives/Science/2017 February 23

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February 23

Does the brain control the heart also?

or the heart is exception and it doesn't control the heart? I'm asking it because according to what I know there are no nerves in the heart. 93.126.88.30 (talk) 00:31, 23 February 2017 (UTC)

See Heart#Nerve_supply. The heart normally beats on its own, but the rate is influenced by the brain. However, trauma to the brain alone can cause cardiac arrest, though our articles do not explain the mechanism. Someguy1221 (talk) 00:38, 23 February 2017 (UTC)

Think you'll find the most important is the Vagus nerve. Beta blockers work by damping down the brains control.--Aspro (talk) 00:41, 23 February 2017 (UTC)

Then the brain controls the heart. Isn't it? Now I'm holding a booklet of the Red Cross organization which states (translated) about the brain that it "activating and supervising on the lungs" while the brain "supervising and regulating the heart". The use different terms for the lungs and for the heart. They don't say about the heart the it's activating by the brain. I'd like to know what is the explanation for these different terms regarding to the brain and the heart. 93.126.88.30 (talk) 01:08, 23 February 2017 (UTC)

Well, it depends on what you mean by control. And brain. The normal function of the heart is primarily controlled by the Cardiac pacemaker without direct continuous input from the brain. But there can be things that happen in your brain that influence the function of your heart. --Jayron32 01:29, 23 February 2017 (UTC)

I think the reason for the difference in description is the nature of the "control". Muscle contractions that lead to breathing are caused by continuous nerve impulses from the brain. That is not the case for the heart, even if the brain can modulate its rate, and the heart cannot beat indefinitely with no input from the brain. Controlled, but not micromanaged? Semi-independent? Someguy1221 (talk) 01:31, 23 February 2017 (UTC)

I'd note that while translated, the word which seems to confuse the OP is "activating". I'd suggest removing "activating" is a fair call for the heart. The heart is supervised and regulated by the brain and does need input from the brain long term, but it isn't really activated by it. It "activates" itself by the cardiac pacemaker which is part of the heart. The lungs do need to be "activated" by the brain, if they don't receive a signal they don't work. The Pre-Bötzinger complex is part of the brainstem, not the lungs. Of course control and regulation of these organs is complicated, as with most things, we you should always be wary of reading too much into words whether English or some other language and with no disrespect to the Red Cross, I'm sure it isn't intended to be a scientific text book. Nil Einne (talk) 01:56, 23 February 2017 (UTC)

Exactly. The difference between the lungs and heart is that the heart contains its own pacemaker, and the lungs do not. No brain = no breathing, but the heart will beat autonomously... for a very brief time. - Nunh-huh 02:19, 23 February 2017 (UTC)

A very rough analogy would be the difference between driving a car and riding a bike - in the car, you control the rate the engine turns by pressing the accelerator, but don't directly affect the details of the motion (analogous to the brain's control of the heart - it can change the rate, but the heart controls the details of the movement). On a bike, your feet are turning the wheels, and there is a 1-to-1 correspondence between the movement of your foot and the movement of the bike (assuming fixed gears - this is like the brain's control of the lungs, where it will send a signal for each breath in and out). MChesterMC (talk) 10:33, 23 February 2017 (UTC)

• The heart does not require a signal from the brain to function on a basic (as opposed to regulated) level, see SA Node and Fibers of Purkinje. The lungs will entirely fail to function without a brain signal. This is called a "central apnea since the failure originates in the central nervous system. μηδείς (talk) 03:51, 23 February 2017 (UTC)

The autonomic nervous system controls heart rate.
Sleigh (talk) 19:42, 23 February 2017 (UTC)

Sudden unexpected death in epilepsy (SUDEP) is a recognized clinical event, though its causes aren't well-understood at present - but it's definitely a case where the brain controls the heart. "Cardiac arrest associated with epileptic seizures: A case report with simultaneous EEG and ECG", Fatemeh Fadaie, et al, Epilepsy & Behavior Case Reports Volume 2, 2014, Pages 145–151 reports on two patients who experienced asystole after epileptic seizures. In the 1970s, I witnessed what may have been a similar case, a young lady who presented at the hospital emergency room where I worked (as a non-medical employee) experiencing what appeared to be drug-related seizure activity which ended with cardiac failure, after which cardioversion failed and she died. loupgarous (talk) 21:58, 23 February 2017 (UTC)

Isn't the brain also involved here by sending signals such that in the absence of the signal the heart rate would rapidly increase to about 100 bpm? I remember reading somewhere that this allows for rapid stress responses. Count Iblis (talk) 22:28, 23 February 2017 (UTC)

Our article on the vagus nerve's function explains that. A properly functioning right vagus nerve does down-regulate the heart beat by controlling the sino-atrial node. The vagus nerves and spinal ganglionic nerve keep the heart rate down to normal levels in healthy people. When the vagus nerves are hyperstimulated bradyarrhythmias, pathologically slow heart beat, can result. Hyperstimulation of the left vagus nerve can cause conduction block at the sino-atrial node.

One correct answer to the OP is that in healthy people the brain does routinely help regulate heart rate through the parasympathetic nervous system - though even after a vagotomy the heart continues beating, still controlled by its internal pacemakers. The control, however, doesn't respond to external stimuli as responsively as when the vagus nerves are left intact. loupgarous (talk) 22:57, 23 February 2017 (UTC)

Ratio of circumference to diameter

Could the ratio of a circle's circumference to its diameter be possibly different in an alternate universe?

Then it wouldn't be a circle. --DHeyward (talk) 09:27, 23 February 2017 (UTC)

• (edit conflict) Everything is possible in an unspecified alternate universe, so yes.

Outside the realm of speculation, you can try reading about Non-Euclidean geometry, where such a ratio is not even constant. For instance, a circle on the Earth's surface of small dimensions compared to the Earth's curvature will have about the same properties as in planar geometry, whereas a circle on the equator has a perimeter of four times its radius (= the pole-equator distance) (assuming a spherical Earth, which is not exactly true), so that ratio is 4 rather than 2π. Tigraan^{Click here to contact me} 09:28, 23 February 2017 (UTC)

"Pi is approximately 3.1415927, given Euclidean geometry" appears to be a true statement regardless of the nature of the universe. If we lived in a universe that was obviously non-Euclidean, we would still be able to imagine a world that was not. As Tigraan mentions, certain non-Euclidean worlds give non-constant values of pi (pi here being not mathematical pi, but the circumference-to-diameter ratio of an arbitrary real circle). You could even imagine a universe in which particle movement was restricted to triangular or square grid lines on a physically realized spacetime, and wind up with pi of exactly 3 or 4, respectively. But then we could still imagine a Euclidean world, and calculate pi ~ 3.1415927. In logic this is called a logical truth, a statement that (given its underlying assumptions) appears to be true no matter what (in the words of philosophers, in every possible universe). Whether logical truths really exist, and what they mean, has been debated for millennia. Though most people just go about their day and don't worry about it. Someguy1221 (talk) 09:44, 23 February 2017 (UTC)

In Pratchett's Going Postal there´s a machine in which pi = 3. The inventor was really annoyed that pi was so "messy". As a consequence, the machine bends time and space. Gråbergs Gråa Sång (talk) 10:24, 23 February 2017 (UTC)

The USA tried that too: Indiana Pi Bill Andy Dingley (talk) 14:45, 23 February 2017 (UTC)

• It's different on this planet too.

Pi's value doesn't depend on the universe, but rather the geometry. If you have a non-Euclidean geometry, then pi doesn't have the same value one would expect for a flat plane. Nor do the angles of a triangle add up to 180º.

You can demonstrate this with a large spherical ball (easy to find), or the alternate of a hyperbolic surface (a bit harder to find physical examples of - a Pringle or a trumpet bell are sometimes used to make museum displays). With an exercise ball, some whiteboard pens and flexible tape measure and protractor you can do classroom demonstrations with measurements. As the radius is measured along the curved plane of the ball's surface it is "longer" than you might expect for a circle of such diameter, thus the value of pi is smaller than for a flat Euclidean plane. Andy Dingley (talk) 10:47, 23 February 2017 (UTC)

Minor nitpick: that supposes you define π as the ratio of circumference to diameter. Many people define it as the half-period of the cosine function, or a similar definition from calculus; you can prove that the cosine, defined as a power series, has a period without involving geometric arguments. See Pi#Definition. Tigraan^{Click here to contact me} 13:12, 23 February 2017 (UTC)

Well, that still leaves the same basic question, but worded differently "Under what mathematics systems does the half-period of the cosine function NOT equal the ratio of the diameter to the circumference of the circle?" The symbol used to represent those two concepts is identical under the mathematics we all know and love, but that's merely convention. What if they were different numbers? --Jayron32 13:15, 23 February 2017 (UTC)

The broader question is whether mathematics (so maybe try the Maths desk) works the same in an alternate universe, a question that has been asked here many times, most recently here.--Shantavira|^{feed me} 12:27, 23 February 2017 (UTC)

• The OP's question amounts to the tautology "If math were different would math be different." The simple answer is "of course it would, you've already stipulated that it was different". The more interesting questions come from asking "If math were different in this specific way how would that one change propagate through the entire system to change other things." Entire fields of mathematics are dedicated to answering that question. Besides the alternative geometries mentioned above in non-Euclidian geometry there are things like Alternative algebra or Non-associative algebra or Non-standard model of arithmetic the like in which some fundamental axiom of a mathematical system is changed, and then further implications are studied. --Jayron32 13:05, 23 February 2017 (UTC)

Newly discovered earth-size planets at nearby star 40 light years away.

This cnn.com[1] page states: "Standing on the surface of one of the planets, you would receive 200 times less light than you get from the sun, but you would still receive just as much energy to keep you warm since the star is so close." It does not make any sense to me. --AboutFace 22 (talk) 13:21, 23 February 2017 (UTC)

Compare a candle to a bonfire. To get the same heat energy from the candle as from the bonfire, you'd have to stand MUCH closer to the candle than you would the bonfire. --Jayron32 13:22, 23 February 2017 (UTC)

(EC with below) I imagine the OP's question is why you receive much less light but just as much energy (since both are referring to a planet that is very close). The CNN story doesn't explain very well, but I imagine they are talking about visible light. Since TRAPPIST-1 is an ultra-cool dwarf, the energy in what's visible light to us would be significantly reduced even on a relative basis. Of course any life evolved there would be adapted to that spectrum. Nil Einne (talk) 13:46, 23 February 2017 (UTC)

• There are multiple effects at play here, and the article seems to be mixing them up awkwardly.

Obviously, the closest from the star you are, the more energy flux (by surface area) you receive. However, that energy is transferred by radiation, and is thus distributed among the radiation spectrum according to Planck's law. This law involves temperature in two ways:

1. The hotter a body, the more radiation it emits overall (see Stefan-Boltzmann law).
2. The hotter a body, the more the bulk of the radiation is shifted to the short ("blue") wavelengths (see Wien's displacement law).

The important point is that "light" refers to the portion of energy that falls within the visible spectrum, where the Sun emits most of its radiation (which is not coincidential: life forms evolved optical sensors called "eyes" to have sensitivity in the domain where there was stuff to see).

My guess is that the star discussed in the article is cooler than the Sun, thus emitting less energy overall and having a peak emission at longer wavelengths (in the infrared domain) than the Sun. Because it is closer, it compensates for the first effect (overall energy flux), but it still emits more at non-light wavelengths. Tigraan^{Click here to contact me} 13:42, 23 February 2017 (UTC)

 

Is it possible that the article really meant that the star was 200 times less bright than the Sun? They actually wrote "200 times less light", but maybe they were just confused. You could get comparable amounts of light from a much dimmer star, if its disk is much larger in the sky.

This issue made for a difficult problem to solve at the Moon article, which claims that the Moon is second only to the Sun in terms of brightness as visible from Earth. Literally, that's false (Venus is about 10x brighter, as you can easily see when they're close to one another in the sky). But the article text goes on to explain that this is as measured by illuminance. I'm not entirely happy with the solution there but I'm not sure I have a better suggestion. --Trovatore (talk) 23:17, 23 February 2017 (UTC)

The stellar magnitude of Venus can be as much as -4.6, and that of the moon -13. The sun's is -27. A unit decrease in magnitude represents an increase in brightness of approximately 2.5. The absolute magnitude of stars is calibrated by lining them up at a distance of 32.6 light - years - for planets and asteroids the distance is a more useful one earth - orbit radius. 80.5.88.48 (talk) 08:49, 24 February 2017 (UTC)

You're missing the point. The Moon delivers more light to the Earth; there's no question about that. But Venus is brighter. Brightness is the amount of light per unit solid angle as seen by an observer, and Venus subtends a much smaller solid angle. (Relative) magnitude doesn't measure brightness, but rather illuminance.

This sounds abstract, but it really isn't. Next time Venus is close to the Moon on a clear night, just look at them. You can easily see that Venus is brighter. --Trovatore (talk) 09:03, 24 February 2017 (UTC)

sub-Q - Is CO2 a greenhouse gas?

I gotta ask... a "Venus-like atmosphere" has not been ruled out for the inner planets of the star. But if they are heated almost entirely by infrared radiation, is a Venus-like CO2-intense atmosphere a greenhouse gas for them, or does it actually make them cooler, or does it more or less cancel out? Wnt (talk) 13:46, 23 February 2017 (UTC)

Can you explain more why you think it might not be a "greenhouse gas", due to the spectrum of radiant heat? Somewhat relevant: you may enjoy this [2] critique of the word greenhouse gas, as it is a terribly misleading metaphor. An actual greenhouse acts to keep the inside warm by limiting convective transport. An actual greenhouse would keep the inside warmer than the outside, no matter what the spectrum of light. SemanticMantis (talk) 15:59, 23 February 2017 (UTC)

The thought is correct (if the star emits in the infrared, the CO2 will absorb more of the incoming radiation), but even a red dwarf has an effective surface temperature of 2300K to 3800K, while the planet would have a temperature about an order of magnitude lower. Wien's displacement law tells us that a planet at 300K will have its maximal emissions at 9.6μ, while a red dwarf at 3000K will have the maximum at 0.97μ. CO2 mainly absorbs at 2-3μ, around 4μ, and from 8-20~13-~16μ. So while the greenhouse effect would be a bit weaker than for earth, it would still contribute to warming the planet, not to cooling it. --Stephan Schulz (talk) 16:37, 23 February 2017 (UTC)

An aside on real greenhouses, and how they work. Not terribly relevant to either question above, interesting though it may be. SemanticMantis (talk) 17:08, 23 February 2017 (UTC)

My understanding is that in a physical greenhouse the windows allow the visible and ultraviolet to get inside, the latter warm the plants and the soil. The latter emit low frequency infrared light which the windows do not allow to pass, thus the energy is trapped inside. --AboutFace 22 (talk) 16:21, 23 February 2017 (UTC) That effect is tertiary at best. Actual greenhouses keep the inside warm by keeping that air in one place, while allowing light in. This is is explained more clearly by a Dr. Fraser at the link I posted above. If you could magically make the walls of a greenhouse permeable to air while keeping the exact same optical properties, it wouldn't be able to keep the inside warm in the winter. Likewise, if you contain a volume of air, butlet the IR back out, you get pretty much the same warming, as illustrated by the quote below. SemanticMantis (talk) 16:33, 23 February 2017 (UTC) (ec) One way an actual greenhouse works is that the glass is almost 100% transparent to visible light, but less so to infrared. Therefore, light enters, hits things like plants, is converted to infrared, but then reflects off the glass and stays inside. If the same is true of the greenhouse effect, then this Q would come up, if there's little visible light. StuRat (talk) 16:23, 23 February 2017 (UTC) Stu, if you'd read what you yourself just linked, you'd see that's not the relevant part at all. It's also explained in my link, and I also explained why greenhouses and the greenhouse effect are not very similar. Since you can't be bothered to read the article, Let me help you and and do some of your homework for you: “ However, R. W. Wood in 1909 constructed two greenhouses, one with glass as the transparent material, and the other with panes of rock salt, which is transparent to infrared. The two greenhouses warmed to similar temperatures, suggesting that an actual greenhouse is warmer not because of the "greenhouse effect", but by preventing convective cooling, not allowing warmed air to escape. ” Next time read and think a bit before you type up whatever you happen to recall from your distant school days. SemanticMantis (talk) 16:31, 23 February 2017 (UTC) Far am I from defending Stu (usually), but the greenhouse effect, while not the primary explanation for the warming of greenhouses, is measurable and even significant - see Greenhouse_effect#Real_greenhouses and Greenhouse#Design. Wood's experiment is interesting, but was rather crude by today's standards. --Stephan Schulz (talk) 16:42, 23 February 2017 (UTC) Sure it's measurable, but significance is... debatable at best. Do you have any sources showing it as contributing to warming in a real greenhouse in anything over small percentages? Because 8% is an interesting footnote, not the thing that we explain if we only get to explain one thing. Saying a greenhouse keeps the interior warm due to the greenhouse effect is like saying "an airplane flies by pushing air down with its wings, and that pushes the airplane up". Yeah, we have Ground_effect_(aerodynamics), but the offered explanation is still very bad and misleading, even though some aircraft can take advantage of this downward push in some circumstances. When we explain things, we should explain the bits that are important, not the bits that are not important. Anyway, I'm sorry I brought it up at all, none of this is terribly relevant to OP or Wnt's question. SemanticMantis (talk) 17:05, 23 February 2017 (UTC) SemanticMantis, I said "One way an actual greenhouse works..", not that this is the only way, or even the main way. The source I provided backs up what I said, and I said that to explain the term "greenhouse effect", since it refers to that portion of the way a greenhouse works. You owe me an apology for misrepresenting what I said. StuRat (talk) 18:16, 23 February 2017 (UTC)

• Convenience link: CO2 transmittance in the IR range. For the answer to the question, see Stephan Schulz's post, although I believe their "8-20μ" indication of an absorption window in the far infrared to be erroneous (going by the spectrum I linked, it is more like 13-17μm). And yes, in theory, the "greenhouse" effect could lead to cooling rather than heating. Tigraan^{Click here to contact me} 17:10, 23 February 2017 (UTC)

Also see anti-greenhouse effect. --Stephan Schulz (talk) 17:16, 23 February 2017 (UTC)

And you are right about the absorption - I was going by a bad web source. This is better. Note that this is maintained by the NIST, but they still claim copyright. That sounds doubly wrong to me... --Stephan Schulz (talk) 20:54, 23 February 2017 (UTC)

A very interesting and well-referenced post that was not asked for by anyone above. If you would like to read why it is unlikely that you will get to travel to one of these planets, and find it a hospitable world of easy wealth, read on.

Reality check A science fiction artist has been at work painting "What the TRAPPIST-1 planetary system may look like" of which 2-3 of the 7 planets resemble our own watery blue marble. This imaging is entirely imaginary, in the tradition of the sci-fi illustrator Chesley Bonestell 1888 - 1986. All that can be said reliably is that existence of 7 probable Exoplanets has been deduced from obscurations of a little star in Aquarius (constellation) designated TRAPPIST-1, 3 of them may be in "Goldilocks" habitable orbits, which is a factor that the speculative Drake equation uses to estimate the chances of Extraterrestrial life, but the reality is that we as yet are less able to image the TRAPPIST-1 planets properly than Sciaparelli was in 1877 to draw the Mars that he could actually see. So to you eager futurists who are already planning to terraform a TRAPPIST-1 planet, already counting the wealth in untouched natural resources to be extracted, already imagining yourselves as colonists of an extraterrestrial utopia, already imagining exquisite sunset scenes where nobody can say for sure whether there are atmospheres or that the planets are spherical, and even planning to build a Dyson sphere or two, I offer the following reality check on your dream: Suppose a long-lost relation leaves you $1 million with which you may buy the fastest car possible - it must be a model in production (but the budget allows you to tune it for higher performance). Now, disregarding little problems like refuelling, let's suppose you drive to TRAPPIST-1 at the car's full speed. Please answer here below 1) your car model, 2) your speed, and 3) your age when you finally arrive at TRAPPIST-1. Blooteuth (talk) 20:46, 23 February 2017 (UTC) EXTRATERRESTRIAL TRAVELS TO OTHER STARS ARE IMPOSSIBLE. That car of yours will take millions of years to get there. --AboutFace 22 (talk) 21:36, 23 February 2017 (UTC) Only under current technology. ←Baseball Bugs What's up, Doc? carrots→ 21:51, 23 February 2017 (UTC) There will be no technology for interstellar travels in a million years. You are up against the laws of physics in there. --AboutFace 22 (talk) 22:26, 23 February 2017 (UTC) We don't necessarily know all the laws of physics, nor all the possible ways to get from point A to point B. ←Baseball Bugs What's up, Doc? carrots→ 22:32, 23 February 2017 (UTC) I think the ref desk has work within the known parameters of physics as we understand it. It's all very well to claim we can be reborn as unicorns or we can travel at 8 million times the speed of light or Donald Trump isn't orange, but most of that is just stuff of fiction right now. Answering questions at the Ref Desk assumes that you're going to respond within the known known parameters, not the known unknown parameters. The Rambling Man (talk) 22:45, 23 February 2017 (UTC) The poster claimed interstellar travel is impossible. I don't think there is universal agreement on that. Keep in mind that in 1902 the poster might have said that about airplanes. ←Baseball Bugs What's up, Doc? carrots→ 23:14, 23 February 2017 (UTC) If you need to send a file to someone, you attach it in your email and send it electronically. You don't need to book a flight and bring a computer with you that has the file on its hard drive. Similarly, I think that advanced civilizations will have made the transition to a machine civilization who travel electronically by uploading themselves to machines at the destination via radio communications (or perhaps using lasers as they'll have a larger bandwidth ). That way you can travel at the speed of light without needing a spacecraft. This only works as long as there are suitable machines at the point of destination, so different civilizations may travel to each other, but travel to locations without technology will be more difficult. Count Iblis (talk) 22:56, 23 February 2017 (UTC) Why are you talking about art? Did someone ask about art? Why are you talking about terraforming? Pretty sure that has not been asked about here either. Can you explain to me why this post of yours, in a whole new section, is anything other than WP:SOAP? We are WP:NOTAFORUM. I don't really disagree with anything you said, but I know this is a reference desk, and not a place to post our tangential grievances with how some people may misinterpret news of scientific findings. SemanticMantis (talk) 23:08, 23 February 2017 (UTC) My launchpad and first link is this cnn report introduced by the OP. Have you noticed how much of it is artwork? Have you noticed the amount of artwork that enhances our TRAPPIST-1 article? It even has a video simulation. Those who remember the Golden Age of Science Fiction (you may not) understand the importance of conjectural art as the only way we (excepting the odd astro- or cosmonaut) will ever perceive things in space. But without a clear idea of what is fact and fiction, the artwork used in NASA's announcement can blind some to the fact that today's status of extrasolar exploration is comparable to that of the makers of Early world maps who decorated unknown lands (which in 1566 meant most of North America) with art to say "Here be dragons". NASA have good reason to popularize their work because education is the only answer to misinformed accusations of criminal, nay "lunatic" conspiracy that somehow persist among Americans who need to be persuaded by this better informed opinion: "Right now, NASA’s annual budget is half a penny on your tax dollar. For twice that—a penny on a dollar—we can transform the country from a sullen, dispirited nation, weary of economic struggle, to one where it has reclaimed its 20th century birthright to dream of tomorrow."[3]. If there are aliens with a Radio telescope on a TRAPPIST-1 planet who follow Earth affairs, they are getting news of a new US President (named Carter), the test flight of Space Shuttle Enterprise, an air collision on Tenerife and the opening of George Lucas's movie Star Wars; on audio they hear the US debut of Pavarotti and the last concerts of The Supremes and Elvis Presley; and the aliens may be delighted to receive news of the Wow! signal event, and may know who sent it. Blooteuth (talk) 02:38, 24 February 2017 (UTC) Ok. That's all fine. I would not have balked at or collapsed a brief comment clarifying that the OP's link contained a lot of speculative art. The thing is, nobody asked for an essay on the value of art in science communication, or to comment on their fantasies. Yet that's what you're doing here. If you have a problem with NASA, take it up with them, but nobody here was even talking about flying to these planets or what they look like. OP and the rest of us were talking about the actual science and physics of exoplanets. I understand your points, I am somewhat sympathetic, and you really have provided some nice references. But I think you may be missing the point, which is that this is not a place to post unsolicited essays. That's called soap-boxing, and while it's good fun in real life and in some internet fora, we have explicit rules against it here. If you and Wnt want to keep talking about it, that's fine, I'll leave you to it. SemanticMantis (talk) 22:42, 24 February 2017 (UTC) Realistically, it is at least conceivable to make a Voyager-like probe that leaves the Solar system on a slow trajectory toward these worlds. If we could make a sufficiently durable probe, it would be possible that it could be loaded with certain very long-lived microorganisms with intent of dispersing them on arrival, polluting those planets and changing them. (Caveat: I don't know about the relative velocity of that system, and if it is too much all these ideas get ever more unlikely) This might permit Earth style life to endure, for better or worse, for a thousand times longer than it would otherwise. So it is not, even with near-present technology, particularly unrealistic to think about them. Now as for looking at them - we're getting better telescopes all the time, and their atmospheres are part of the transits of the stars. It may be very difficult but it is by no means impossible to figure out what compounds are in them, their approximate temperature, maybe even look for so-called signatures of life. And then... there are the very brave or very foolhardy people who like to send radio signals at things beyond our comprehension. Wnt (talk) 11:45, 24 February 2017 (UTC) Foolhardy yes, because like Lucy we may have some 'splainin' to do. The awesome introduction to "Contact" (video) plays the babble we have already transmitted in a sphere of radius 123 Light-years (counting from Guglielmo Marconi's transmissions in 1894) that is still expanding. See History of radio. Blooteuth (talk) 15:02, 24 February 2017 (UTC)

Can a human house provide shelter for the animals, and in return, the animals provide food for humans?

Spiders and insects can somehow come into the human-constructed house. Termites can chew on the wood. Rats can enter the house through the toilet or a rat hole in the structure. Then, there are the intentional animals, like cats and dogs and chickens. Some humans live in a rural area, so chickens can enter the house and get on the furniture. While cats and dogs and chickens probably cannot survive on their own, other animals can enter and leave at will. It seems that the human house functions like a little world. If the house provides shelter for these animals and attract animals because of food, then can humans take advantage of that and use the house to lure animals in instead of going after the animals in another place? 66.213.29.17 (talk) 16:58, 23 February 2017 (UTC)

Sure. Just leave the doors and windows open, and shoot anything edible that comes in. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:06, 23 February 2017 (UTC)

What sorts of references were you looking for? I'm unclear as to what additional information you want us to provide for you. --Jayron32 17:14, 23 February 2017 (UTC)

Yes, the human house is a little world. You might be interested in reading about Commensalism. Things like house centipede and house fly house mouse and flour beetle basically only exist in human-made environments. Flour beetles are edible, but usually considered a pest. Our article on entomophagy lists some other human commensal species, and if you'd like to learn more about bug eating, see this excellent article from the FAO [4], which describes how eating bugs can increase our food security. Some people do eat mice [5]. If you try attracting mice to your house to eat, let us know how that goes! SemanticMantis (talk) 17:19, 23 February 2017 (UTC)

• A traditional Swiss chalet, and many other vernacular architectures, was a form of farmhouse where dairy animals were kept inside the same building for at least part of the year. They would also need to be fed: this was done by fattening them through the Summer on good pastures outside, then keeping them indoors over Winter on stored hay or silage. They food they provided though would mostly be in the form of cheese - this was produced from the animals during the good feeding season, then could be stored for consumption during the hard Winter. Some animals (not just under this Swiss model) would also be consumed at Christmas feasts, in the earlier part of the hard Winter, and could then be either eaten at a last feast, removed from needing to be fed through the Winter, or could be preserved as hams, sausages etc.

The problem comes down to a food supply: many farming societies have kept animals in the same house during Winter, but they've not been able to feed them there, other than from stored food. I can't think of clear examples where the animals are kept "in the house" year-round, as there's no way to produce their food in-house too. Chickens or birds in a dovecot might be "indoors" year round, but they'd need to be fed on high-value cereals too.

Could you eat termites? Only if they're eating your house too. If I eat spiders, how do I get the flies to feed my spiders? My houseflies disappear in the Winter. Andy Dingley (talk) 17:24, 23 February 2017 (UTC)

Yeah, you run into basic energetics problems with closed system that's too small. Metabolic_theory_of_ecology may be of interest to anyone pursing such plans. (My desktop salticids seem happy year-round, and never have to leave throughout the winter. They don't need to eat that often. If I were concerned, I'd just leave some old bananas around ;) SemanticMantis (talk) 17:38, 23 February 2017 (UTC)

• My grandfather used to keep livestock in the basement, which served as pets for my mother and her siblings, until holiday dinner. μηδείς (talk) 22:49, 23 February 2017 (UTC)

I wonder if a set of beehives can be built around a small house, providing free heat and well-nigh unbeatable security. ;) But no one I know of has much experience training guard bees, you'd need broad vents for the summer, and you probably have to make your whole property a flower garden. Still, there must be a place to put one in a sufficiently whimsical work of fiction... Wnt (talk) 11:58, 24 February 2017 (UTC)

• This is a quite common system, called "rented flat". Nowadays, most of the animals pay cash or even by money transfer, but paying in kind is not unknown. --Stephan Schulz (talk) 12:54, 24 February 2017 (UTC)

They do. --IEditEncyclopedia (talk) 15:27, 24 February 2017 (UTC)

Intensive farming can be of many forms. But for the OP's " It seems that the human house functions like a little world." there is a problem - these intensive farming practices require food to be brought into the "house" from outside. For dairy, this is often high value food too, grown and manufactured at some cost. Andy Dingley (talk) 16:05, 24 February 2017 (UTC)

Kinetic energy, Momentum

Is there a Conservation of kinetic energy?

For example: An object $m_{1}=2kg$ moving at a constant velocity $3{\frac {m}{s}}$ , elastically collides with a resting object $m_{2}=1kg$ .

Let us say now that $m_{1}$ is resting and $m_{2}$ is moving at a constant velocity $6{\frac {m}{s}}$ . We get

$m_{1}{\vec {v}}_{1}=m_{2}{\vec {v}}_{2}\ \Rightarrow \ 2\cdot 3=1\cdot 6$

but

${\frac {m_{1}v_{1}^{2}}{2}}={\frac {m_{2}v_{2}^{2}}{2}}\ \not \Rightarrow \ {\frac {2\cdot 3^{2}}{2}}={\frac {1\cdot 6^{2}}{2}}$

This is impossible. Where was I wrong? יהודה שמחה ולדמן (talk) 17:53, 23 February 2017 (UTC)

What is your basis for concluding that it's wrong? Akld guy (talk) 18:18, 23 February 2017 (UTC)

You assumed the final velocities are 0 and 6 m/s. As you found above, those assumptions are not consistent with an elastic collision. Dragons flight (talk) 18:26, 23 February 2017 (UTC)

Looking at elastic collision might help. Dragons flight (talk) 18:29, 23 February 2017 (UTC)

I still don't get it. יהודה שמחה ולדמן (talk) 18:42, 23 February 2017 (UTC)

• Why would you think that ${\vec {v}}_{2}>>{\vec {v}}_{1}$ ?

We don't know ${\vec {v}}_{2}$ and it's hard to predict, without understanding details of the collision, such as the elasticity of the masses. This isn't trivial in theoretical kinematics, it's near impossible for practical experiments, except by making empirical measurements. But we do know that momentum is conserved and that the KE must be reduced: it can't increase, it can't even stay constant except for a theoretical and perfectly elastic collision. So we can put an upper limit on ${\vec {v}}_{2}$

${\frac {m_{1}v_{1}^{2}}{2}}>{\frac {m_{2}v_{2}^{2}}{2}}\ \Rightarrow \ {\frac {2\cdot 3^{2}}{2}}>{\frac {1\cdot v_{2}^{2}}{2}}$

${\vec {v}}_{2}<{\sqrt {2\cdot 3^{2}}}\approx 4.24$

Andy Dingley (talk) 19:05, 23 February 2017 (UTC)

But for the conservation of momentum we get that $v_{2}=6{\frac {m}{s}}$ . So why does the kinetic energy smaller? יהודה שמחה ולדמן (talk) 19:11, 23 February 2017 (UTC)

Sorry, can't typeset that quickly!

My ${\vec {v}}_{2}$ limit above is for the case you describe where $m_{1}$ is left at rest afterwards. What this proves is that $m_{1}$ cannot be at rest afterwards, from this collision. You can't set up any physical case where all momentum is transferred and the initial mass stops dead. Andy Dingley (talk) 19:15, 23 February 2017 (UTC)

You have two boundary conditions, which you can now solve for a possible physical case: momentum is conserved (it's conserved as a principle), energy is conserved (it need not be, but that establishes a boundary) but there is no axiom that one mass ends up stationary, and we've just shown that for masses like these it's not even possible for it to do so.

To solve this, and establish both ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$ after the collision, solve the simultaneous equations that you have for momentum and energy. But don't assume ${\vec {v}}_{1}=0$ Andy Dingley (talk) 19:18, 23 February 2017 (UTC)

I think Andy Dingley has covered it, but I had to write it out a different way to really see it. Maybe this will help OP as well.

One problem is that you (the OP) have overconstrained the system by specifying both an elastic collision and the final velocities. By specifying an elastic condition, you have assumed that kinetic energy is conserved. In this case, use energy equivalence (and momentum equivalence) to calculate the final velocities. If instead you specify the final velocity, use this to calculate the final kinetic energy.

A second problem is that the final velocity that you have specified results in an increase in kinetic energy of the system, which is impossible without an additional input of energy. That some final velocities for m1 are impossible is a valuable result of the thought experiment IMHO.

A third, and more minor, problem is notation. Since you have both initial and final velocities for m1 and m2, labeling velocities as simply v1 and v2 can be confusing. Using v1i, v1f, etc. might make things clearer-Wikimedes (talk) 01:54, 24 February 2017 (UTC)

In case it was unclear, $v_{2}=6{\frac {m}{s}}$ with $v_{1}=0{\frac {m}{s}}$ is one way to conserve momentum, but there are an infinite number of ways to conserve momentum. For any choice of v2, you can find a corresponding v1 that allows the total final momentum to be equal to the initial momentum. An elastic collision means finding the values for v1 and v2 that allow both momentum and kinetic energy to be conserved. Dragons flight (talk) 08:38, 24 February 2017 (UTC)

Let's pick a convenient frame of reference: the one in which the total momentum is zero. This is the centre-of-mass frame (CM-frame). As the mass ratio of $m_{1}:m_{2}$ is 2:1, the speed ratio in the CM-frame must be 1:2, so ${\vec {v}}_{1}=1m/s$ and ${\vec {v}}_{2}=-2m/s$, the CM-frame itself moving at ${\vec {v}}_{CM}=2m/s$. In case of an elastic collision, conservation of both energy and momentum dictates there are only two solutions: the initial velocities, or both velocities reversed. As the masses don't move through each other without interacting, we know that after the collision ${\vec {v}}_{1}=-1m/s$ and ${\vec {v}}_{2}=2m/s$. Converting to the original frame of reference, we see that ${\vec {v}}_{1}=1m/s$ and ${\vec {v}}_{2}=4m/s$. PiusImpavidus (talk) 10:01, 24 February 2017 (UTC)

To directly answer the boldfaced question "Is there a conservation of kinetic energy": No there isn't. Energy can be readily converted from one to another form. For example, if a ball is launched straight up, kinetic energy is created from some other form of energy in the launcher (chemical potential, mechanical potential, electromagnetic, whatever); as it slows and stops under gravity, the kinetic energy is converted to gravitational potential energy (and a little is converted to heat, through friction with the air); when it falls again, the gravitational potential energy is converted back into kinetic energy; and when it hits the ground, the remaining kinetic energy is converted into heat and sound energy. --76.71.6.254 (talk) 21:00, 24 February 2017 (UTC)

Alright, to start off with I'm not that clear on your collision calculation. An object at rest doesn't have kinetic energy in that frame... maybe one is before and one after ... I didn't follow it. Anyway, the easiest way to solve these I think is to go into a frame where the net momentum of the colliding objects is zero. So 2 kg and 1 kg collide with one another while converging at 3 m/s. The rate of convergence doesn't depend on frame unless you want to go relativistic, and right now you're not ready for warp speed. So to make the momentum equal we need 2 kg to approach the center of gravity at 1 m/s and 1 kg to approach it at 2 m/s; both have 1*2 kg m/s momentum. They bounce off each other elastically so they carry away equal energy, and their total momentum doesn't change either, so we know they move away from each other at 1 m/s and 2 m/s the same way they came in. Now if we shift the frame so that the 1 kg object started at "rest" (adding + 2 m/s to the right, assuming it was on the right), it ends up with a speed of 4 m/s, while the 2 kg object starts at 1 + 2 = 3 m/s (as you said) and ends at -1 + 2 = 1 m/s. So the kinetic energy for the two at the beginning in this frame was 0.5(2*3^2 + 1*0^2) = 9, and at the end it's 0.5(2*1^2 + 1*4^2) = 9. The kinetic energy is indeed conserved in an elastic collision when looked at in any one inertial frame of reference, assuming the objects involved aren't doing work against (or being accelerated by) any other sort of force like gravity or friction. However, viewed in another frame it will be different, just as every object at rest can be viewed in a frame where it is moving. Wnt (talk) 22:25, 24 February 2017 (UTC)

The solution with 1 m/s and 4 m/s assumes one dimension, i.e. that after the collision the objects are moving along the same line as before. This doesn't have to be the case. See Elastic collision#Two-dimensional. The speeds can become anything from 1 to 3 m/s for the large object, and 0 to 4 m/s for the small. They need 2v₁² + v₂² = 18 (m/s)² to conserve kinetic energy, and there is no solution conserving momentum if v₁ < 1 m/s or or v₂ > 4 m/s. PrimeHunter (talk) 00:34, 25 February 2017 (UTC)

Quantifying change in equilibrium compositions in gas phase reactions from halving/doubling reaction volume

Hi, I just have to double check this. If I halve the partial pressures in the Haber process gas phase equilibrium constant, do I end up square rooting the equilibrium constant? But how wouldn't this apply in say, the solution phase for acid-base equilibria? (Changing reaction volume should affect gas phase reactions but not solution phase.) Yanping Nora Soong (talk) 22:10, 23 February 2017 (UTC)

Note I am trying to quantify the change not merely note the direction. Yanping Nora Soong (talk) 22:11, 23 February 2017 (UTC)

The answer to your first question is "no"; the equilibrium constant is a constant (unless changing the volume or pressure causes a temperature change as well, or your gases are far enough from ideal that you need to take fugacity into account). The change in equilibrium concentrations or partial pressures caused by a change in initial pressures is a different matter, and I would recommend looking up "equilibrium calculations" in a freshman chemistry textbook to see how this is done. It's not hard, but it wasn't obvious to me how to do it before I saw it done.--Wikimedes (talk) 01:21, 24 February 2017 (UTC)

OK, looking at Haber process I see that the equilibrium constants K_p we give are based on partial pressure of the gases, unlike the more familiar K_c based on concentration, but they work the same way. There is a molar fraction x_A of each gas, and the partial pressure of each is that times the total pressure in the container. Then as with the concentration variant, the concentration is equal to the concentrations of all the products over all the reactants, where multiple molecules count multiple times, so it's the square of the ammonia concentration over the nitrogen multiplied by the cube of the hydrogen. The constant stays the same ... unless, as our table shows, the temperature is changed, because then the entropy term changes while the enthalpy stays the same. Wnt (talk) 22:09, 24 February 2017 (UTC)