Wikipedia:Reference desk/Archives/Science/2012 July 9

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July 9

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EQUATION HELP;;;

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pro e can plot curvs with equations we give. I am in urgent need of the following equation.Any mathematician kindly help me..:


i need equation for a twisted pair wire(2 wires only wound in form of a helix along a circle.Such as a two

TWISTED CABLE PICTURE

The figure below shows a twisted cable .i need equation for such a twisted curve made along a helix. as such a twisted wire in form of a telephone cable.RIGHT!!

TELEPHONE CABLE

Sameerdubey.sbp (talk) —Preceding undated comment added 03:26, 9 July 2012 (UTC)[reply]

Your 2nd link doesn't work for me. If you simply want a 2D model, use a positive and negative sine wave for the two wires. For example:
Y = sin(X)/10
Y = -sin(X)/10
I divided by 10 to flatten them out a bit, you can change this if you want them flattened more or less. Let us know if you need to plot a 3D helix, and/or add thickness to the wires, or if you need to simulate 3D by having the wire in front alternate each time they cross. (However, beware that each of these changes make the implementation more complex.) StuRat (talk) 03:37, 9 July 2012 (UTC)[reply]


Thanx a lot for quick reply.However i want a 3d model.and the wires would not overlap,they are twisted actually as shown in the figure.I think this would become simpler if i think in terms of cylindrical coordinate. i forgot to mention that the helix must form a closed circle.


i am giving a link for twisted wire once again. telephoner cable

once again stating my whole proble. 1.a twisted wire(2 strands only) 2.in form of helix(like telephone cable) 3.helix must form a closed curve

actually i am trying to make 3d model for a circular core wound with twisted wire.ok plz help 203.197.246.3 (talk) —Preceding undated comment added 06:37, 9 July 2012 (UTC)[reply]

Do you want something like this: [1], but modeled in 3D ? Also, you didn't say whether you need to represent wire thicknesses, or just show the center-lines. If you do want to show the thickness, do you want a surfaced model or a solid model ? StuRat (talk) 06:51, 9 July 2012 (UTC)[reply]
If you just want the center-lines, it's just (cos kt, sin kt, t) and (-cos kt, -sin kt, t), where k is a constant proportional to how twisted you want it. If you want thickness, you could approximate it with (cos ku + r cos v, sin ku + r cos v, u) where r is the minor radius if it's not very twisted, or with (cos ku (1 + r sin v), sin ku (1 + r sin v), u + r cos v) if it is. If you want something more accurate, I'd have to think about it more, so I'd rather not until I know that's what you actually want. You could toy with that last r to make it look better. — DanielLC 20:11, 9 July 2012 (UTC)[reply]

Silicone dioxide in vitamins for hair

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I have read sbout silicone dioxide in food as additives what about hair vitamins? I am researching a product sold on the internet under a company called Hair Essentials for Women. This product is being marketed to women for thinning hair or want stronger thicker hair> My concern is how safe is it in capsule form? Does it really affect the linning of our stomach? — Preceding unsigned comment added by 174.112.156.3 (talk) 04:02, 9 July 2012 (UTC)[reply]

I've never heard of silicone dioxide, are you sure you have that right ? Silicon dioxide, on the other hand, is basically sand, so I wouldn't expect it to be particularly helpful or harmful (unless inhaled). AFAIK, it's just filler, often added to dilute super sweet artificial sweeteners to get them closer to the net sweetness of sugar. StuRat (talk) 04:17, 9 July 2012 (UTC)[reply]
It is an additive used as said as a filler and to prevent powders from clumping. The danger from ground glass is mechanical. It has to be above a certain size to pose a threat of irritation. The grade used in food (like taco seasoning powder) and drugs is non-problematic. μηδείς (talk) 04:23, 9 July 2012 (UTC)[reply]
Agreed, unless it's inhaled. I also would think it would tend to grind down the teeth over the years, since it's very hard, but the capsules should avoid this (on the plus side, it might help to remove dental plaque and dental calculus when not in capsules). I also wonder if it qualifies as non-nutritive dietary fiber. StuRat (talk) 04:29, 9 July 2012 (UTC)[reply]
Yes, when I worked as a prep-cook I used to make the chili, and the ground glass would form a nice white cloud no matter how I tried to pour it. See Silicon_dioxide#Health_effects. Baking soda and salt, or even table sugar is much better if you need an abrasive to brush your teeth. No, it is definitely not fiber, fiber is absorptive. You want to keep your intake to a minimum: the poison is in the dose. μηδείς (talk) 04:49, 9 July 2012 (UTC)[reply]
I had a similar problem with cinnamon forming a cloud, but figured out a solution. I pour it directly into liquid (like applesauce), which limits the cloud size. Then, before I turn the container back upright, I put the lid on. This prevents the second cloud that comes out when the cinnamon falls back down to the bottom of the container. I try to do all this quickly, while holding my breath, and then evacuate the area, to avoid breathing any cinnamon clouds that were produced. (Anyone who has ever inhaled a cloud of cinnamon powder will appreciate why I take such precautions.) StuRat (talk) 00:29, 10 July 2012 (UTC)[reply]

Why can't the extra dimensions of M-theory be temporal?

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M-theory requires 7 hidden dimensions, so why must these be spatial rather than temporal dimensions? Hcobb (talk) 12:25, 9 July 2012 (UTC)[reply]

I don't know much about this, but the importance of 9+1 dimensions in string theory is related to a special property of Spin(9,1) (see [2] at the end of the page). That property isn't shared by Spin(8,2), Spin(7,3), etc., though it is shared by Spin(8) and, I think, Spin(10,2) and so on. M-theory adds one spatial dimension to that. Don't ask me why. -- BenRG (talk) 17:41, 9 July 2012 (UTC)[reply]

environmental science

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what are the different types of biomes? — Preceding unsigned comment added by Scmcforpeace83 (talkcontribs) 12:46, 9 July 2012 (UTC)[reply]

That seems like something that could be answered by reading our article entitled "biome". DMacks (talk) 12:49, 9 July 2012 (UTC)[reply]

Newton's atoms in my body

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Now there's a wide spread saying that your body contains atoms that were once in [put a famous person here]'s body. Now that seems nice, and the reason they offer for this is that atoms circulate in nature and since the number of atoms in human body is huge, there are at least some atoms in our body that belong to somebody. Well, first of all, I agree that elements circulate in nature, but how long does it take for them to do so? given that some of the people they exemplify are pretty recent in history, I'm skeptical about this claim, I mean I'm sure that atoms in my body were there in the body of some prehistoric animals for sure, but individuals in recent history?! and given all the plants and animals (including people) that have ever existed, aren't there at least some individuals that didn't make it to my body? It seems a bit unlikely to me. does the "shuffling" of atoms through history happen this fast and this radically?--Irrational number (talk) 13:24, 9 July 2012 (UTC)[reply]

Water circulates much faster than solids, and air even faster than that. It is certainly plausible that most of the air you breathe contains atoms that have been around the world in relatively short timeframes. I'm not so sure how well the claim holds up in the end though. Even though we have ridiculously huge numbers of atoms in us, there are a lot more atoms in the environment that aren't currently part of any human. Just because you have something like 10^25 (for example - i made the number up) more atoms than the number of people that ever existed doesn't necessarily make it likely that at least one has come from each person. 209.131.76.183 (talk) 14:20, 9 July 2012 (UTC)[reply]
(ec)Molecules in air move as fast as the wind, recent examples of which are radioactive fallout from the Chernobyl disaster in Ukraine arriving 36 hours later in Finland and the rapidly changing area of Air travel disruption after the 2010 Eyjafjallajökull eruption. You don't have to be on this planet very long to have a high probability of sharing gas with him and her. DriveByWire (talk) 14:40, 9 July 2012 (UTC)[reply]
We did the math for this in high school. The result is pretty convincing, although we assumed for simplicity that once in the air a molecule stays in the air, and that people don't rebreathe the same molecules, and came up with an estimate of 1,000 atoms per breath having been breathed by Jesus as well. Even correcting for air molecules leaving the atmosphere and people not breathing unique molecules in every breath the number should remain well over 100. The math desk might work out a formula for you. μηδείς (talk) 17:04, 9 July 2012 (UTC)[reply]
Saying that one specific person's atoms are likely in your body is VERY different from saying that the atoms of every individual who's ever lived is in your body. Suppose that, for a random person, you're 99.999% likely to have one of his atoms. That means that for 1 million random individuals, there's only a 1 in 22,000 chance (0.99999^1000000) that you'd have an atom from every single person. --140.180.5.169 (talk) 17:32, 9 July 2012 (UTC)[reply]
Yes, but the likelihood is much, much, much, much, much much higher than 99.999%, which is merely 1/100,000. How many atoms are there in a mole, how many moles does someone breathe a day, and how many days do they live? The number that gives is incredible. Divide it by the number of liters in the atmosphere and you get molecules per liter. A helpful visualization to comprehend the numbers is to imagine a raindrop magnified to the size of the earth. The water atoms would be the size of basketballs. That's just one drop of water. μηδείς (talk) 17:40, 9 July 2012 (UTC)[reply]
I'll point out, as I do every time this topic comes up, that atoms and molecules are indistinguishable quantum particles and hence it's quite meaningless to ask whether a particular atom now is the same as a particular atom from a long time ago. It probably does make sense to talk about what fraction of yourself was formerly Newton, but it doesn't make sense to compute a probability that you have at least one former Newton atom, because all atoms are a mixture of former Newton and non-former-Newton. -- BenRG (talk) 17:56, 9 July 2012 (UTC)[reply]
You realize, of course, that your interpretation of that article ("all atoms are a mixture of former Newton and non-former-Newton") is entirely wrong. Atoms may swap outer electrons, but the electrons themselves and the nuclei of the atoms maintain their integrity excepting radioactive decay. μηδείς (talk) 18:27, 9 July 2012 (UTC)[reply]
It's not an interpretation of that article, but a true statement about quantum mechanics. If the article really says otherwise then it should be fixed. The reason that particles lose their identity when in close proximity is that every interaction has contributions from Feynman diagrams in which the particles switch places and other diagrams in which they don't. -- BenRG (talk) 19:06, 9 July 2012 (UTC)[reply]
The central point of that article is that electrons, protons, and neutrons are indistinguishable from each other because they don't have smaller identifying features. Any electron is as good as another. But under normal circumstances, nuclei simply do not flow into one another. That's Heraclitus or Deepak Chopra, not Richard Feynmann. μηδείς (talk) 19:22, 9 July 2012 (UTC)[reply]
If that's the point of the article, it's misleading. That's why we can't tell them apart, not why we know they're the same. We know they're the same because quantum states with the particles switched interfere, which only happens if they're the same state. For example, if the amplitude of state A is 0.1 and the amplitude of state B is 0.1, the probability of getting A is 0.1^2 = 0.01, and the same for B. The probability of A or B is 0.02. If the two states are the same except with two identical bosons switched, the probability of getting either state is 0.04. If they're not the same state, that's a 0.01 chance of getting A, a 0.01 chance of getting B, and a 0.02 chance of ... what exactly?
You could claim that the universe is just symmetric in such a way that any two particles have the same/opposite amplitude when switched. If you accept the Many Worlds interpretation, this would mean that you share the atoms of an equally likely alternate universe Newton. If you accept the Copenhagen interpretation, it means that any time two identical particles get entangled, they have a 50% chance of switching. Quantum entanglement has been found on relatively large scales, which would seem to imply that any action on a smaller scale would cause entanglement. The quarks in nearby atoms are constantly interacting via., among other things, their electric charges. Every time the waveform collapses, the quarks could end up in any of the nuclei. — DanielLC 19:51, 9 July 2012 (UTC)[reply]
It seems that the original poster has asked a specific question, and gotten some very ambiguous answers shrouded in debate about the appropriateness of various methodologies. ..."How long does it take" ... for a molecule to move from place to place... this is predominantly dictated by the mean free path and the drift velocity; for molecules in common substances, an answer can be determined, subject to realistic parameters. The extent to which the calculations of those parameters will require quantum-mechanics, as opposed to being suitably-well approximated by classical statistics, will depend on the type of molecule. So, for example, a very large organic carbon compound will probably behave fairly well like a classical particle (albeit one with a strange shape, moment of rotational inertia, and so forth); while an individual oxygen or hydrogen atom will probably behave in a way that is much better described using the ugly statistical formulations that quantum physics requires to accurately make predictions. For lack of a better model, and given that the OP hasn't precluded the use of a spherical cow model, it seems fair to assume that every molecule on the planet (including those inside living organic tissues) could be considered as a member particle of a gigantic Maxwellian gas, whose temperature is approximately the average surface temperature of our planet; so simply calculate the average molecular speed, and you've got an approximation for the rate of particle circulation. You can use the diffusion equation to estimate the mean distance traveled by any individual particle. Once you've got that model working, you can add in as much additional complexity as you like to correct for all of the terrible, invalid assumptions that are clearly due to such a simplistic model. If you're interested in classical statistical physics, I always recommend the red book of Statistical Physics and Thermodynamics. If you're interested in quantum mechanical statistics, there exists a plethora of very bad texts to choose from. Nimur (talk) 21:48, 9 July 2012 (UTC)[reply]
"We are star dust."
"Yes, but we're also made of dino poo." :-) StuRat (talk) 18:56, 9 July 2012 (UTC) [reply]
So we're both golden ... and brown? Clarityfiend (talk) 20:50, 9 July 2012 (UTC)[reply]
...and we've got to get our dino poo back to the garden. StuRat (talk) 21:00, 9 July 2012 (UTC) [reply]
Back to the quantum physics, can we please get a ref on how often atmospheric nuclei in different molecules at room temperature (or even the temperature of lava) can be expected actually to intermingle? In atoms per year? As for the mean free path, it is something like 10cm, travelling at 1,000kmph, at room temperature, but that is hardly relevant on the timescale since newton. Weather mixes the atmosphere pretty well on that sort of time scale. μηδείς (talk) 22:14, 9 July 2012 (UTC)[reply]
If you're using quantum physics, the answer is that they're already the same nuclei. You could mess around with the question enough to get something that looks like the sort of answer you'd expect (so long as the particles don't get too close together), but the question itself is fundamentally flawed.
If you insist on messing with it to get the sort of answer you're looking for, the atoms in your body and the atmosphere intermingle every time you eat and breathe. When you digest food, the carbon in it gets stored in fat. When you exercise, the carbon in your fat gets exhaled in CO2. The CO2 gets absorbed by plants, and gets eaten by someone else. I'm not sure exactly how long it takes to get spread around in the air, but the bottleneck is probably the time between plants getting picked and eaten. — DanielLC 05:09, 10 July 2012 (UTC)[reply]
My thinking as well; even ignoring the rather crucial issue of unique persistent identity we could still spend days just proposing metabolic, climatological, geological, and ecological factors that need to be taken into account, to say nothing of resolving them with any precision. That being said, the OP is clearly looking for a ballpark estimate -- that is, is it more likely than not? And my (half-)impressionistic take on it is that he has reason to be skeptical - the likelihood of possessing an atom from any one particular historical figure (again, assuming we ignore the indescernability principle), especially a recent figure, is probably small - if not exactly astronomically small. That's a rather unempirical answer, I know and I hope to see the math take root here for something more concrete, but I'd be surprised if even rough estimates suggest it to be likely. If we're going to try though, might I suggest A subpage? since the discussion is likely to get quite involved? Unless someone knows a reason not to create a subpage on the RefDesk. I can always copy our work over to subpage in my user space later if we aren't finished by the time this thread is archived. Snow (talk) 12:11, 10 July 2012 (UTC)[reply]
Medeis, I just want to mention that the identity question I'm talking about comes from whole nuclei swapping places, not from any process involving the constituent nucleons. The amplitude for some kind of mix-and-match rearrangement of two nuclei is vanishingly small because of the energy barrier, but there's no energy barrier to this nucleus going over there and that nucleus coming over here. They just go around each other, they don't pass through each other or teleport. Nevertheless they follow Bose-Einstein or Fermi-Dirac statistics in these processes. If they merely had the same properties (but weren't "indistinguishable" in the quantum sense) they would follow classical statistics and it would be possible to figure out which was which after an arbitrary amount of time. -- BenRG (talk) 16:30, 10 July 2012 (UTC)[reply]
So now that atoms are maintaining their identity? Before it was "all atoms are a mixture of former Newton and non-former-Newton". You have yet to make a coherent claim and provide a ref for it. μηδείς (talk) 16:39, 10 July 2012 (UTC)[reply]
Medeis, with due respect, I think you need to stop attacking BenRG's words (because, more often than not, he tends to be exactly correct); and instead, you should reacquaint yourself with exactly which quantities are known to be quantized. Those quanta describe physical parameters that are defined as "indistinguishable" according to the indistinguishability principle - because they have no other properties. Nimur (talk) 18:05, 10 July 2012 (UTC)[reply]
I never claimed that atoms of the same isotope are distinguishable based on their properties. Frankly, if BenRG wants to come here and imply how tired he is of having to correct people all the time the responsibility certainly does lie with him to make a coherent and consistent statements and to back his facts with refs. To say theoretically that two identical atoms may switch places with each other with and unspecified frequency in no way indicates that it is meaningless to say that one is breathing atomic nuclei which someone else breathed. It's an absurd claim based on a new agey wagey pseudo-grasp of science. I don't intend to comment on this any more, since it might imply I think the matter is open to argument. But I will be impressed when I see numbers and refs. μηδείς (talk) 22:52, 10 July 2012 (UTC)[reply]
I hate people like Chopra who say "quantum mechanics!" to justify whatever crap they believe in. And I don't like to call quantum mechanics weird or unintuitive because I think that's a cop out. But there is a reason that many physicists describe it that way. What I'm describing, whether it sounds strange or not, is an essential and basic feature of quantum mechanics. I don't understand what you think quantum mechanics is, if you don't think it's this. The only reference that makes sense is any introductory quantum mechanics textbook, or maybe QED (book). In QED there are Feynman diagrams like the ones I was talking about in figures 59 and 60 (pages 93 and 94) in my copy. Please don't come back and say that those diagrams don't prove anything. I'm not saying that the diagrams by themselves prove anything. I'm just trying to point you to a relevant part of the book, in case you don't want to read the whole thing.
I'm sorry I sounded testy in my first post. I was kind of testy at the time, but that's no one's fault but mine. -- BenRG (talk) 05:14, 11 July 2012 (UTC)[reply]
The reference desk has had a similar question more than once but I couldn't find it in the archives. Here is a similar off wiki discussion: [3]. Rmhermen (talk) 15:42, 10 July 2012 (UTC)[reply]

The fact that atoms are identical in QM is actually a trivial issue here (the state is a sum over permutations of the labels you assign to the atoms). What you need to do is formulate the problem in terms of a well defined correlation function. Count Iblis (talk) 17:47, 11 July 2012 (UTC)[reply]

environmental science

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describe chaparral. give and describe the three types of chaparral. differentiate those three

S.C.M.C."WARFARE" (talk) 13:28, 9 July 2012 (UTC)S.C.M.C."WARFARE"[reply]

We have an article on chaparral. Read it. Then do your own homework. AndyTheGrump (talk) 13:31, 9 July 2012 (UTC)[reply]

A friend recently visited Cook Inlet, Alaska where she saw the locally endangered Beluga. Why, she asks, is it still legal to use their eggs for caviar if they are endangered? μηδείς (talk) 17:13, 9 July 2012 (UTC)[reply]

Whales are eutherian mammals, and thus do not lay eggs. Beluga caviar is the eggs of the beluga (sturgeon), which lives in the Caspian Sea, and not in Alaska. AlexTiefling (talk) 17:16, 9 July 2012 (UTC)[reply]
According to beluga (sturgeon), this fish is "critically endangered", so I suppose the question still makes sense... -- Ed (Edgar181) 17:27, 9 July 2012 (UTC)[reply]
In terms of a worldwide ban, the article also suggests the obvious answer is correct, i.e international politics. Unfortunately (IMO) it is generally difficult to get action to protect some endangered species when a fair number of countries, particularly influental ones are against the idea. However the OP's comments are unclear. As far as I know they live in the US and their friend was obviously visiting a part of the US. According to our article and supported by [4], importing Beluga caviar has been illegal in the US since 2005 and illegal to trade in interstate commerce. It's still legal to trade intrastate it if it was imported before the ban, but it seems likely any beluga caviar that remains legally imported in the US must now be very rare and therefore very expensive whatever the state (although there is probably some variance). If the OP or their friend is aware of a source of Beluga caviar in the US that isn't so I suggest they contact the appropriate authority (FWS?) as it's likely it's illegally imported. BTW, the earlier source which despite some later statements only mentions specific bans on Beluga caviar from the Caspian and Black Sea may lead one to believe it's possible to get it from the Adriatic Sea which our article mentions as a possible habitat. But this apparently incredibly outdated source (it doesn't mention the 2005 US bans nor does it mention the 2006 CITES ban although it does support the idea international politics are making a ban difficult) [5] suggests they haven't been seen there since 1980. Nil Einne (talk) 19:00, 9 July 2012 (UTC)[reply]
What I wonder about is whether the sturgeon will be fished to extinction, or whether the fishing will stop once the price gets too expensive for the consumers to justify. ←Baseball Bugs What's up, Doc? carrots09:15, 10 July 2012 (UTC)[reply]
It seems to be a veblen good, so maybe not. StuRat (talk) 16:47, 10 July 2012 (UTC)[reply]
The reason the beluga whale and the beluga sturgeon share the same word in their name is that both derive from the Russian word белуха (belukha), meaning "white". Ghmyrtle (talk) 09:40, 10 July 2012 (UTC)[reply]
Beluga caviar is not white. μηδείς (talk) 22:43, 10 July 2012 (UTC)[reply]
No, the but the fish is white(ish). 203.27.72.5 (talk) 23:16, 10 July 2012 (UTC)[reply]

Terrell rotation matrix

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How do you find the matrix for Terrell rotation? I understand that Terrell rotation isn't quite linear, but I just want the first approximation. — DanielLC 19:58, 9 July 2012 (UTC)[reply]

Just apply the Lorentz transform to the nearest- and farthest- points on the object, or to any vertices you like; and fit a rotation to the transformed vertices. (If you only use two vertices, that is very straightforward). This is covered in Tipler's introductory relativity chapters, if I recall correctly; and does not require very advanced mathematics. Nimur (talk) 21:31, 9 July 2012 (UTC)[reply]
No. You'd have to calculate when and where the points crossed your past time cone. Since it's nonlinear, it would only work in the limit. Of course, the answer itself will only work in the limit for the same reason, so that's not that big a problem. You'd need four points to do it since it's not really a rotation (three if you're only doing it in two dimensions, which I am). Finally, I need an equation for it. I was hoping someone, somewhere already calculated it out. — DanielLC 04:47, 10 July 2012 (UTC)[reply]
The easiest way to understand the appearance of a "moving" object is by working in a frame where the object is at rest. Then it's just an ordinary linear perspective projection modified by aberration. I'm not sure how to write down a formula for this—I think it's easier to just understand the geometry of the situation, so you can solve any particular problem of this kind, than to memorize a complicated general formula. -- BenRG (talk) 06:00, 10 July 2012 (UTC)[reply]
I don't want to memorize it. I want to program it. — DanielLC 07:21, 10 July 2012 (UTC)[reply]
Well, what I said still applies—the appropriate formula will depend on what sort of program you want to write. If you want to visually simulate relativistic motion through a static scene, you can do that as follows: do an ordinary nonrelativistic 3D rendering of the scene onto a cube map with the camera at the player's instantaneous location, then, for each point in your final image, take a ray in that direction, modify it by the aberration formula (using the player's velocity relative to the scene), intersect it with the cube map, optionally apply some kind of color and brightness transformation, and you're done. This is the same basic approach as Fisheye Quake except that you're using the aberration formula instead of a fisheye projection. On any graphics card with programmable pixel shaders you should be able to do this in real time for an arbitrarily moving (and accelerating) player.
The relativistic aberration page only tells me how to mess with the angle. Not the distance. Come to think of it, do I do the distance the light was sent from, or would it mess with parallax and end up being closer/further than it appears? — DanielLC 20:17, 10 July 2012 (UTC)[reply]
There's no distance in aberration. It acts on rays of light, not points in spacetime. I assume you're asking because you need a value for the z buffer. If so, the easiest way is probably to pick a reference frame (such as the rest frame of the environment) and use the distance (=time) in that frame. See below. -- BenRG (talk) 02:55, 11 July 2012 (UTC)[reply]
If you're thinking more along the lines of a text-based program that solves a parametrized word problem, tell me what the word problem is and I'll come up with a formula. -- BenRG (talk) 07:53, 10 July 2012 (UTC)[reply]

My program is a relativistic shoot-em-up. It can find where it's supposed to draw a bullet or enemy, but I want to know how to warp the image properly. In principle, I could just have it check where to draw every vertex, and then attach lines from there, but this seems like a waste of processing power. — DanielLC 20:17, 10 July 2012 (UTC)[reply]

I approve of this project. :-) When I first saw Fisheye Quake I thought about doing a relativistic Quake, but never actually got around to it.
I recommend that you render each independently moving object in the way I suggested above: first render to a cube map, then apply aberration (and perhaps other effects) with a pixel shader. Here's some detail about how to do that.
  • You need to keep track of a coordinate frame for each rigid object in the game (landscape, player, opponent, bullets) and a fixed background frame in which you express the dynamical position of everything. You need to know how to transform between all of these. The transformations will vary with time. All this is standard in 3D graphics, but there are two differences here. First, these are spacetime coordinate systems (x, y, z, t), and the transformations between them are Poincaré transformations (Lorentz transformations, rotations, and translations). Second, because light has a finite speed, you need to keep a historical record of coordinate transformations for each object instead of just the "current" transformation. This is complicated, but I'll assume you know how to do it. The rest is pretty easy. :-)
  • To render a frame, do the following for each object other than the player:
  • Find the point of intersection of the player's past light cone with the object's worldline. You mentioned that you already know how to do this. From now on when I talk about object coordinates I mean the object's historical coordinate frame at that point on its worldline.
  • Transform the player's spacetime position into object coordinates.
  • Render six views of the object into six square offscreen RGBZ buffers, with a 90° field of view, with the camera located at the spatial part of the transformed player position and facing in the ±x, ±y, ±z directions. This is standard 3D stuff, except that doing the lighting correctly is tricky. I'm going to ignore lighting for now.
  • For each pixel on the screen:
  • Find the pixel's location in player coordinates. I assume you know how to find the x, y, z coordinates. The t coordinate needs to be −sqrt(x²+y²+z²), so that the line from the player (origin) through the pixel location is a past light ray.
  • Transform the player location and the pixel location into object coordinates. Find the intersection of the (spatial part of the) ray through the two of them with the cube map.
  • Optionally alter the RGB value from the cube map. I'm going to ignore this for now.
  • Find the point on the ray with t = Z, where Z is the Z value from the cube map (I'm treating the Z values as negative integers with 0 being the camera's location). Note that this point lies on the surface of the object. Transform the point into player coordinates (or background coordinates, or any other coordinate system you like, as long as it's the same for every object in this frame). Take the transformed t coordinate as your new Z value.
  • Write the transformed RGB value and transformed Z value to the output buffer.
That's it. Note that I didn't use the aberration formula anywhere in this. All transformations are Lorentz (Poincaré) transformations. I think this makes things simpler (and faster).
If you just want the rotation, not the aberration, as you originally requested, then you can bypass the cube map and write directly to the output buffer. However I don't think it's any easier to do it this way, and it may not even be much faster (on a modern GPU). If you really want to do it, I'll work out some steps.
If you're interested in proper lighting, redshift, etc., I'll try to figure out some way of doing that. It's not easy, and may make the game hard to play.
Note that to really be relativistically correct you need to deal with the twin effect somehow. If player Alice is standing still and player Bob is circling around her, you need to freeze Bob's game occasionally to allow Alice to "catch up". It's not hard to do this correctly, but it might detract from the fun of the game. Alternatively you could slow Bob's game down, but that feels wrong to me, since it gives the wrong impression about Bob's subjective perception of time. -- BenRG (talk) 02:55, 11 July 2012 (UTC)[reply]
That's alright, Bob won't find that the handicap worries him too much. Girls are terrible at first-person shooters. 203.27.72.5 (talk) 05:35, 11 July 2012 (UTC)[reply]

A shoot 'em up is 2D, and generally single player. I am considering making a multiplayer option, mostly just because I figured out how. Shoot 'em ups commonly have a button to slow time down. I could make it so that you can slow time down until you reach your opponent's past time cone, and if you reach their future time cone, it just turns on automatically. This system would also neatly avoid network lag problems.

I suppose I probably have more than enough computing power to check each vertex individually. It's just that transforming the sprites seems simpler and more elegant. Also, that way I won't have to deal with tachyons having parts that you can't see, and other parts that you can see twice. I could limit it to slower-than-light particles only, which would get me out of all the necessary stuff to make tachyons dodgeable, not to mention the time travel paradoxes, but that would just feel like a missed moment of awesome. Especially because of the paradoxes.

I can't imagine leaving out the lighting effects. Imagine all the cool stuff you could do with them. Give a bullet a transparent infrared skull on a violet background, and it will look like a red skull when it's flying towards you, and then turn blue once you dodge it. Paint flowers on an enemy, and you can see those ultraviolet patterns on them when they fly away. Anyway, I've largely figured out how to do it. I'll just have the brightness of each color component be the brightness of the color times the visibility of that frequency at that color receptor. — DanielLC 05:35, 11 July 2012 (UTC)[reply]

In a 2D shooter the camera is effectively at infinity, which makes it of limited use to the player if you take the speed of light limit seriously. If you ignore that problem, then moving objects are simply Lorentz contracted. There's no Terrell rotation in this situation (though the Lorentz contraction can lead to an effective rotation: a line with slope 1, horizontally contracted by a factor of 2, is a line with slope 2). There's also no aberration. The redshift factor is the gamma factor of the moving object (independent of direction). There's never a blueshift. This would probably be kind of boring.
The best I can come up with for this type of game is to have the camera hovering a relatively short distance above the game field, tracking the player (perhaps simultaneously with respect to the lab frame), and filming the field with a wide-angle lens. If the game objects are 3D meshes (even though the action is 2D) then the method I described above will still work, and it will handle everything including Terrell rotation, but with a large number of independent moving objects it may be too slow. I can work out how to optimize it if this is indeed what you want to do. Lighting is still complicated.
The difficulty with Doppler shift is that there's no way to do it correctly in the RGB color model. An object colored #000000 might glow in the infrared, in which case it turns red when blueshifted, or it might not. If the shifts are small then you can probably fake it convincingly. You may also want to think about the brightening/dimming associated with aberration (I don't know if it has a name, but it's related to the headlight effect). -- BenRG (talk) 18:07, 11 July 2012 (UTC)[reply]
I suppose what I'm doing would best be described as taking the point of view from the ship, and then, without correcting for optical abberations, working out what it would look like from above. Think of it like those things they show radar images on that show how far away stuff is and in what direction as if it was a picture taken from above, even thought it was taken from right there. Also, I am not planning on using an RGB color model. I'm planning on using a (frequency, intensity) model.
A radar-like screen can show whatever it wants. Realistically it would show whatever would allow the pilot to shoot things most effectively. It certainly wouldn't show Doppler shift, for example, since that wouldn't be helpful at all. There is no "relativistically correct" thing for it to show, unless you literally want it to use its sensors to figure out what the scene would look like from above, in which case the correct thing to show is what the scene actually looks like from above. Relativity can tell you what you'll see, but it can't tell you what your point of view looks like from someone else's point of view. That doesn't make any sense. -- BenRG (talk) 21:50, 11 July 2012 (UTC)[reply]

Is [6] and Rotation matrix too simplistic? DriveByWire (talk) 15:38, 11 July 2012 (UTC)[reply]

I'm not sure what to say about the external page. It makes things much more difficult than they need to be by doing everything in the rest frame of the camera, instead of the rest frame of the object. I'm not sure how a Lorentz contraction could be visible in this two-dimensional world, where the image in the camera is just a line. The line in the camera does get shorter when the object is moving, because of the headlight effect. The interesting observation that Penrose and others made circa 1960 is that a moving sphere always has a circular outline, not (as you might expect) an oval, but you need all three spatial dimensions for that effect. Regarding the rotation matrix, the easiest way to understand the Terrell rotation is as a real rotation related to the camera's spatial location in the rest frame of the object, so a rotation matrix will work when drawing the object just as in ordinary 3D graphics. But the camera image (not the object itself) is also altered by aberration. If the object has a large angular size at the camera's location then aberration will distort straight lines into curved lines, but if the object is small then aberration will stretch or shrink the image but not distort it much. -- BenRG (talk) 18:07, 11 July 2012 (UTC)[reply]
It is false. The contraction part of Terrell rotation does not counter out Lorentz contraction, and doesn't even necessarily apply in the same direction. If an object is moving towards you, it will look longer than it really is. If it's moving away, it will look much shorter. If you want to know what it really ends up looking like, I suggest checking out Light Speed!. — DanielLC 18:41, 11 July 2012 (UTC)[reply]
Note that the description of optical aberration at http://lightspeed.sourceforge.net/about.html is wrong, which may be the reason for some of your confusion. I think that their images are probably relativistically correct, they just have the terminology wrong (and they're doing it the hard way). -- BenRG (talk) 22:31, 11 July 2012 (UTC)[reply]

Higgs boson

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Is it possible to block Higgs bosons or manipulate the Higgs field in some way to reduce another particle's mass? --146.7.96.200 (talk) 21:07, 9 July 2012 (UTC)[reply]

Not at this time, nor for the conceivable future, save via the intervention of Clarkeian "magic". — Lomn 21:12, 9 July 2012 (UTC)[reply]
Is it theoretically possible? --146.7.96.200 (talk) 21:56, 9 July 2012 (UTC)[reply]
Not in any meaningful sense under present theory (see above need to invoke "magic"). — Lomn 23:00, 9 July 2012 (UTC)[reply]
From my limited understanding, it is roughly as theoreticaly possible as a perpetual motion machine. It would need to violate laws which we hold as some of the most certain laws we have. Vespine (talk) 01:20, 10 July 2012 (UTC)[reply]

Yes, and in the early universe the Higgs field was different, it was zero on average, causing the mass of particles to be zero. When the universe cooled, the Higgs field settled down at its minimum energy, which is not at zero field strength, causing particles to gain mass. Count Iblis (talk) 01:28, 10 July 2012 (UTC)[reply]

Hold on a second. If you heat a system above the electroweak unification temperature, you can no longer describe the Higgs interaction as "giving mass to the particles", but that doesn't mean it just disappears. Unless I'm seriously confused, what you get at high temperatures is a complicated interacting soup that in no way resembles a bunch of free massless particles. Technically they are massless, but technically they're massless at low temperature too.
Also—as you no doubt know but the original poster probably doesn't know—almost all of the mass of ordinary baryonic matter (the stuff we're made of) comes from the strong force binding the protons and neutrons, not from the Higgs field. Even if you could dial the Higgs field down to zero (which you can't) it would hardly make any difference in the mass of any ordinary object. It would make enough of a difference that cellular chemistry would no longer work and everybody would die, but not much of a difference numerically. -- BenRG (talk) 05:45, 10 July 2012 (UTC)[reply]
  • So, do we have an article that explains which particles have mass independent of Higgs, and which would be massless with a zero (high energy) Higgs field present?
  • Also, as I (don't) understand it, the Higgs is a scalar field, so the field level can only be lower or higher than it is now. If I have a majik laser pointer that shoots a big beam of very fast-moving Higgs particles, will the field in the area it lights up (before they decay) be lower or higher? Wnt (talk) 15:53, 10 July 2012 (UTC)[reply]
All of the elementary particles in the standard model, except the Higgs itself, get an effective mass through the Higgs interaction. In composite particles the binding energy and the kinetic energy of the particles also count as mass. So basically it's the elementary particles that are massless.
The Higgs boson is a wave in the Higgs field, so it alternates between higher and lower than normal. -- BenRG (talk) 04:41, 11 July 2012 (UTC)[reply]

And note that you don't need to use the Higgs field to change the mass of particles. What makes the Higgs field special is that it achieves its minimum energy state when the field strength isn't zero. This then causes particles to gain a mass in the vacuum. Then if you want to use some field to make the mass of a particle different from what it is in vacuum, you can use any field that couples to that particle. E.g. the mass of an electron in a very strong electromagnetic field is slightly different from its vacuum value. The shift of the squared mass of an electron in the beam of a laser is given by:

 

where   is the fine structure constant,   the electron Compton wavelength,   the wavelength of the laser beam and   the density of photons in the laser beam. Count Iblis (talk) 01:42, 10 July 2012 (UTC)[reply]

Habitatable zone at the tip of RGB

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About the tip of RGB in 7.59 billion years I was wondering where will earth-like habitable zone be located (Jupiter and Sautrn's system, or more likely Pluto)? General calcualations, yes I know Jupiter and Saturn's orbit will be further out roughly 1.7 times further out (roughly 8.8 AU and 17 AU) but the thing is the luminosity greatly increases by 3000 times. The mathematic calculations yahoo answers figure out is 3000/5.2x1.7 for Europa, and 3000/9.5x1.7 for Titan to figure out the surface condition for Europa and Titan. Is this possible at the tip of RGB, Europa can end up colder than earthlike temperature today of only -40F? I was wondering what is the coldest temperature estimate Europa may be in 7.59 billion in the future. I don't know what other calculations will work according to the tip of RGB, the star's temperature is alot cooler, if that matter.69.226.40.110 (talk) 21:09, 9 July 2012 (UTC)[reply]

I did a quick search and found this mentioned in Terraforming of Europa, alas, not with good sources that I noticed... I see [7] this talks about it reaching Mars, but it says nothing of Jupiter... of course, the huge caveat with all this is that the planetary orbits are likely to shift as the sun expands, and if so... well, just about anything can happen. Wnt (talk) 21:21, 9 July 2012 (UTC)[reply]
If humans (or our descendents) are still around in 7.59 billion years to worry about it, I suspect technology will have advanced to a level where we could just move the Earth wherever we wanted to. Although, by then, only a small portion of humanity may live on Earth (those who like to do things in the quaint, old-fashioned way). StuRat (talk) 21:30, 9 July 2012 (UTC)[reply]

For information, RGB refers to the Red Giant Branch phase of a sun's evolution. That is predicted for our Sun in about 5 billion years. DriveByWire (talk) 21:48, 9 July 2012 (UTC)[reply]

Am I not clear at the first place? I mean when sun peak the luminosity and size at the end of the giant stage at 3000 times the luminosity and when sun reaches the maximum radius of 1.0 or 1.2, could Europa at that time when sun being 3000 times brighter stay at the temperature below -40F? I don't know how to calculate that way. --69.226.40.110 (talk) 22:31, 9 July 2012 (UTC)[reply]
It wasn't clear to me what the B meant, although I knew you were talking about a red giant. StuRat (talk) 23:39, 9 July 2012 (UTC)[reply]
I meant when the sun's giant reaches its maximum size and luminosity could Europa just heat up only up to -40F where Mars surface temperature is today?--69.226.40.110 (talk) 23:46, 9 July 2012 (UTC)[reply]
Pluto will never be a habitable planet. μηδείς (talk) 02:44, 10 July 2012 (UTC)[reply]
As we've explained to you a dozen times before, nobody knows the answers to these questions. You are quoting numbers that are just the best guesses of a particular paper. There are numerous papers, all with different answers, because we're talking about what will happen in billions of years time and there is far too much uncertainty to draw any firm conclusions. Asking essentially the same question again and again is not going to change the answer. --Tango (talk) 14:59, 10 July 2012 (UTC)[reply]
What do you mean by dozen times before? How can you know it was the same person asking? — Preceding unsigned comment added by Plusanother (talkcontribs) 21:56, 10 July 2012 (UTC)[reply]
The same writing style and, essentially, the same question. It is obviously the same person. --Tango (talk) 02:13, 11 July 2012 (UTC)[reply]
To be a lot less rigorous, the habitable zone article points out that this is approximately a simple inverse-square relationship - so to make Jupiter, at 5 AU, inhabitable, you need a Sun 25 times as bright. But looking for more on this, I happened across [8], which suggests that the Sun will get less than 20 times brighter by 12.1 Gyr, at which point it starts pulsing and losing mass catastrophically and they abandoned attempts to model the results. Those authors favored a 200K water-ammonia life possibility on Titan for a few hundred million years. Of course, even 0.01 Gyr is a very long time if you happen to be a good gardener with an extensive assortment of seeds, so I wouldn't rule out the notion of some balmy paradise Jupiter moon existing for some brief span near the end. Wnt (talk) 15:45, 10 July 2012 (UTC)[reply]

Product of animal menstruation?

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Recently I got into a discussion with a radical vegan and she, besides calling me cruel, inhumane, close minded, and so on, also said that I was set on eating products of animal menstruation. Although I did not understand when she said it, I believe she meant eggs. So, first does it make sense? was it really eggs? Is that normal within the vegan community to talk like that about eggs? Second, according to menstruation, it only refers to certain mammals, does any serious person talk about some form of avian menstruation? Third, is there any non-animal menstruation? OsmanRF34 (talk) 22:20, 9 July 2012 (UTC)[reply]

You can't reason with radicals, doesn't really matter what the topic is. You are right that birds don't technically menstruate, but one PART of menstruation: the expulsion of an unfertilized ovum is technically what laying an egg is, so it's not really a massive stretch IMHO to call a chicken egg a "chicken's period". But so what? My fav breakfast is chicken periods with slices of pig's arse, yum! It's an Appeal to emotion, we associate menstruation with "disgust", I personally don't know why, I think it has religious roots really, but I think in fact, there's nothing disgusting about it. In fact, mayube you could ask the vegan what she finds so disgusting about menstruation? Vespine (talk) 22:46, 9 July 2012 (UTC)[reply]
Slight technicality there Vespine, it is not pig's arse, but pig's chest. Pigs arse is still good though!! Caesar's Daddy (talk) 07:30, 10 July 2012 (UTC)[reply]
Many vegetarians are fine with milk and eggs, as no animal has to die to get those. As for your third Q, if you are asking if there are animals that don't menstruate, then yes, there are. For example, cats don't, but dogs do.
A retort you could make, if she likes honey, is to tell her that it's bee spit. Or, if she likes fruit, tell her she is eating plant ovaries. StuRat (talk) 22:53, 9 July 2012 (UTC)[reply]
Bee vomit, actually. Delicious! Ratbone58.164.228.40 (talk) 02:25, 10 July 2012 (UTC)[reply]
She was a vegan: so no animal products (which come from animal slavery, with or without killing). Veganism is all about animal rights in a very radical way.
The third point is about the redundancy: why animal menstruation? Is there something that could be called vegetable menstruation somewhere? OsmanRF34 (talk) 23:25, 9 July 2012 (UTC)[reply]
As far as "animal slavery", that's a fair characterization of factory farming, but how does she feel about "happy" free-range chickens ? StuRat (talk) 23:38, 9 July 2012 (UTC)[reply]
I don't believe vegans share your definition of slavery = factory farming and I don't believe I'll talk again with her about such topics, so my understanding might be limited. Anyway, as far as I could understand, no captivity of animals is acceptable to her: no milk, no eggs. It's not only the killing. It's the slavery. And she has the equivalent of normal people's opinion on slavery but applied to animals. Mainstream citizens do not accept slavery, even if the master treats the slaves well. So, she doesn't accept animal slavery. OsmanRF34 (talk) 00:10, 10 July 2012 (UTC)[reply]
If somebody wanted to provide me with food, housing, and medical care, and, in return, only asked something from me which I couldn't use anyway, like, say, my urine, I'd sign right up ! StuRat (talk) 19:19, 10 July 2012 (UTC) [reply]
THAT wouldn't be slavery, since someone asked you for something. That's trading. Just imagine the opposite case, if you get imprisoned, get lots of food to make you fat, health care only if it were economically viable, otherwise they let you die, and at the end they would eat your fat ass. — Preceding unsigned comment added by Plusanother (talkcontribs) 22:38, 10 July 2012 (UTC)[reply]
You seem to be talking about factory farming, but I was talking about free-range chickens raised for their eggs, which aren't imprisoned, but are free to come and go as they please. They also aren't fed so much that they get fat, since they aren't for eating, but for laying eggs. And comparing the health care they get with the lack of any in a wild bird, it seems like a good deal. StuRat (talk) 23:55, 10 July 2012 (UTC)[reply]
I would say she's talking out of her arse (if you'll pardon the expression), birds don't menstruate. I think you will find the Menstruation (mammal) article more informative. If you meet her again, tell her she's flogging a dead horse (sorry!). As for non-animal menstruation, it's the shedding of the uterine lining so, by definition, it's only possible in animals. Richerman (talk) 23:32, 9 July 2012 (UTC)[reply]

She means animal as opposed to human. I agree with Vespine's and StuRat's points. My last girlfriend that was a vegetarian would eat eggs and milk, just "nothing with eyes". And she would point out to people that she wore leather because there was no suitable substitute, but would prefer something else if it were available. Much more reasonable than telling people they were eating chicken abortions. μηδείς (talk) 23:48, 9 July 2012 (UTC)[reply]

The funny thing is, the factors that drove the evolution of brains big enough to allow us to worry about these ethical dilemmas are thought by some to include the consumption of fish and shellfish and that we later learned to cook meat. Richerman (talk) 00:01, 10 July 2012 (UTC)[reply]
Just out of curiosity, what use of leather can't be substituted with something else? Wnt (talk) 00:23, 10 July 2012 (UTC)[reply]
I suppose elegant shoes, specially men's ones, have to be of good quality leather. OsmanRF34 (talk) 00:30, 10 July 2012 (UTC)[reply]
Quality hiking shoes were her main concern. She moved to Colorado from the east coast due to her love of winter sports. μηδείς (talk) 00:47, 10 July 2012 (UTC)[reply]
What about substituting hiking for canasta? 203.27.72.5 (talk) 01:25, 10 July 2012 (UTC)[reply]
What in bloody hell would she need leather shoes to play canasta for?
Umm.....if she played canasta instead of hiking, she wouldn't need the shoes, and the cows wouldn't have to die for her hobby. 203.27.72.5 (talk) 03:04, 10 July 2012 (UTC)[reply]
The cattle are dying anyway. And the poor canasta plants? Who's thinking about their well-being? μηδείς (talk) 03:41, 10 July 2012 (UTC)[reply]
Just got to say to those claiming veganism is extemely radical, you need to check out Jainism, they won't even kill plants or bacteria. Unique Ubiquitous (talk) 01:36, 10 July 2012 (UTC)[reply]
They certainly will kill bacteria and various other microorganisms simply by having a functional immune system. 203.27.72.5 (talk) 02:04, 10 July 2012 (UTC)[reply]
I assume their religion only requires them to attempt to minimize the deaths they cause, not eliminate them entirely. I'd also assume they put more value on large animals than bacteria. StuRat (talk) 02:44, 10 July 2012 (UTC)[reply]
The immune system kills invaders, not the native flora, whose cell numbers outnumber those of your body by ten to one. μηδείς (talk) 02:49, 10 July 2012 (UTC)[reply]
Are you absolutely sure that the immune system never, ever kills any of the native flora in the body? I'm pretty sure some opportunistic infections are due to native flora that are normally kept in check by the immune system. 203.27.72.5 (talk) 03:16, 10 July 2012 (UTC)[reply]
Ah, yes. Candida albicans is a good example. 203.27.72.5 (talk) 03:19, 10 July 2012 (UTC)[reply]
It's biology, not math. Of course there are exceptions. Such organisms are like clonal colonies, in any case, not groups of unique zygotic individuals. The body does not usually kill them off entirely. I am not a Jain, but I would assume the relevant point is that you are not voluntarily going out of your way to kill bacteria which inhabit your body without invitation in the first place.
Well, my point is that regardless of what Jains might say or think or believe, they will kill a variety of things no matter what they do. 203.27.72.5 (talk) 03:44, 10 July 2012 (UTC)[reply]
Your italics sound just like my niece. You understand the difference between killing a leaf and killing an entire unique individual plant? Almost all infections are by clones, identical cells, like the leaves of plants rather than individual zygotes. μηδείς (talk) 03:53, 10 July 2012 (UTC)[reply]
I note with no little amusement how the supposedly radical vegan manages to justify using leather. It reminds me of something Mark Twain said, in the voice of Huck Finn talking about his aunt, who was always condemning smoking and drinking. "She took snuff, but that was alright, cause she done it herself." ←Baseball Bugs What's up, Doc? carrots09:12, 10 July 2012 (UTC)[reply]
I think you have confused Osman's radical vegan with Medeis' vegetarian ex. AlexTiefling (talk) 09:40, 10 July 2012 (UTC)[reply]
I was actually thinking of someone I know who's a "devout" vegan who also smokes cigarettes. Go figure. ←Baseball Bugs What's up, Doc? carrots14:58, 10 July 2012 (UTC)[reply]
What's the problem with that? Cigarettes are not animal products.
For most vegans, their health is at least part of the reason for their dietary choices, which makes such an unhealthy choice as smoking surprising. StuRat (talk) 17:16, 10 July 2012 (UTC)[reply]
OK, but there are also enough vegans who are vegans due to ethical reasons. It's not necessarily a contradiction. — Preceding unsigned comment added by 188.76.173.109 (talk) 18:13, 10 July 2012 (UTC)[reply]
Most vegans I know smoke all sorts of things that aren't good for their health, and I really don't finding it suprising at all. According to Hippie; "Hippies rejected established institutions, criticized middle class values, opposed nuclear weapons and the Vietnam War, embraced aspects of Eastern philosophy, championed sexual liberation, were often vegetarian and eco-friendly, promoted the use of psychedelic drugs which they believed expanded one's consciousness, and created intentional communities or communes." 203.27.72.5 (talk) 20:12, 10 July 2012 (UTC)[reply]
It's ironic that their ethics toward others apparently does not extend toward themselves. I wonder, though, if they got their wish and all "enslaved" domestic chickens were turned loose, how long would it be before canines and raptors had pretty well wiped them off the face of the earth? ←Baseball Bugs What's up, Doc? carrots23:03, 10 July 2012 (UTC)[reply]
Actually, feral chickens do quite well for themselves, although they are not caged layers. μηδείς (talk) 23:43, 10 July 2012 (UTC)[reply]
And frankly it's a poor argument for all of the excesses of the meat industry to claim that you really are just concerned with what would happen to all the poor animals if you weren't keeping them in small pens and slaughtering them. It's just not very likely that that's what's keeping you up at night. (I'm not a vegan, but I don't pretend that factory farming is anything ethical.) --Mr.98 (talk) 00:28, 11 July 2012 (UTC)[reply]
There's no question that the extreme factory farming that you're alluding to is cruelty to animals. It's a fact of life that every living thing eats and eventually gets eaten. It's good to treat domestic animals as humanely as possible or practical, up to and including the method of slaughtering. ←Baseball Bugs What's up, Doc? carrots02:58, 11 July 2012 (UTC)[reply]
We're definitely out of our league here. If someone wants to ask how it is that humans can think that we can enforce an ethical code on behalf of animals that the animals themselves have never followed toward one another and never will follow, better ask the folks at the Humanities desk... Wnt (talk) 21:32, 11 July 2012 (UTC)[reply]