Wikipedia:Reference desk/Archives/Science/2009 February 17

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

February 17

New colors...Editing Break

Argh. I wish I'd noticed this thread sooner. Listen up, folks:

• We do not see the world in red, green, and blue.
• Our three cone types are not red, green, and blue, despite the misleading coloring on that diagram above, which I changed a few months ago for this very reason but it got reverted. Here's my version; the colors aren't any more sensible but at least they aren't the exact three colors that are most likely to mislead people.
• To the best of my knowledge the L (long-wavelength) cone does not have a secondary peak in the violet. The diagram I just linked, which shows no such peak, is correct as far as I know. More importantly, though, there's no reason to expect any such peak, once you understand how color vision actually works. If you think that the three cones are red, green, and blue then you might expect such a peak, but they aren't, not in any way, shape, or form, so that whole line of reasoning is irrelevant.

It's easy to understand how color vision actually works, you just have to wipe your mind of this nonsense about red, green, and blue. Ready? Okay. There are three cone types in the eye, each with a different response curve. This means that the infinitely many different wavelengths that make up a physical color spectrum are projected down to just three nonnegative numbers by the cones. What wavelengths is each cone sensitive to? It hardly matters at all, for reasons that I hope will be clear in a minute (though I'd like to again emphasize that the answer is not red, green, and blue). The important thing is that there are three dimensions. One of those three dimensions is just overall brightness. If you divide out by that you're left with two dimensions. One popular coordinate system for that two-dimensional space was standardized by CIE and is imaginatively called the "xy" coordinate system. The diagram at the start of this thread is plotted in xy coordinates.

The following is not understood by the brain's visual system—it is a modern scientific discovery: If you plot the responses of the cones to monochromatic light in xy coordinates, you get the Λ-shaped curve that forms most of the boundary of that diagram. The cone responses are linear (twice the incoming energy with the same frequency distribution → twice the response from each cone type). The xy coordinates are a perspective projection of the linear cone-response space (just like in 3D graphics, with the "eye" at (0,0,0), i.e. perfect blackness). Therefore, given two physical colors (that is, spectra) which map to (x₁,y₁) and (x₂,y₂), any linear combination of the two colors will map to a point on the line joining (x₁,y₁) and (x₂,y₂). Any linear combination with nonnegative coefficients will map to a point on the line segment with (x₁,y₁) and (x₂,y₂) as endpoints. This means that the only (x,y) coordinates you can reach with physical spectra are those in the convex closure of the Λ-shaped curve, which is why that part is colored.

Here's what the brain understands: There's a two-dimensional space of colors with a distinguished white point in the middle. Different directions from the white point are interpreted as different hues, and different distances from the white point are interpreted as different saturations. The brain doesn't know anything about monochromatic light, and it has no reason to care: it can't see the physical spectrum, all it gets is these two coordinates (plus brightness).

So why does violet look like a combination of red and blue? Because it's between red and blue in the only color space the brain knows about. Frequency has nothing to do with it. There's no highest or lowest in the brain's notion of hue—it's a circle.

Getting back to the original question, if you stimulated the cones in a way that's impossible with physical light, what would you see? I can only guess, but presumably you would see supersaturated versions of the same old hues. It would be a new visual experience, but not a new hue, which is what people probably mean by "color" in this context.

Red, green, and blue are used for color reproduction. With three colored phosphors on a display monitor you can reproduce any color in the triangle with the (x,y) coordinates of those phosphors as its vertices. If you look back at the CIE diagram it's pretty clear that the largest triangle you can fit in there will have vertices at red, green, and blue/violet. So those are the phosphors used on real displays—except that they aren't because nothing ever makes sense. For practical reasons real displays use phosphors that cover a rather small triangle on the diagram, but they're still red, green, and blue and they still cover the whole circle of hues, just not the whole range of saturations. These days the primaries are usually sRGB, shown here. That image itself uses sRGB (as does the one at the top of this thread) and your monitor is probably calibrated in sRGB, which means only the colors inside that triangle are correct; the others are not saturated enough. (They're also the wrong hue because whoever made these diagrams clamped the RGB channels individually to [0,1] instead of clamping in the direction of the white point, which makes the green portion look a fair bit larger than it should.) -- BenRG (talk) 01:27, 19 February 2009 (UTC)

BenRG loves to cut in at the end of a long (and largely correct) explanation and claim that everything that's preceeds his remark is stupidly wrong...however, more than 50% of the time that's not true. This time doubly so. OK - so let's take his points individually:

• There is no green sensor. Well - kinda. The actual center frequency of the 'mid-range' sensor is more towards yellow - and the PERCEPTION of green and yellow is much confused by the large amount of overlap with the red sensor. But we CALL it the green sensor...and you aren't going to change that. Green is the color seen by the green sensor...by definition.
• Hence we do see the world in Red, "Green" and Blue...by definition.
• You aren't aware of the 'red bump' up in the middle of blue. Well, yes, it's a relatively new finding (compared to the ancient nature of the original set of curves). However it's the thing that explains the human tendancy to position a color called 'violet' or 'purple' or even 'magenta' between blue and red on a "color wheel" - even though physics tells us that red and blue are the ends of the spectrum. Weird - but true.
• One of the three color 'axes' is intensity - well, we're talking about a 3D 'color' space - and elementary math says that you can represent any point in a 3D volume by three numbers - and you can map the space with spherical polar coordinates, cylindrical polar coordinates, axes at right angles, axes that aren't at right angles...lots of possibilities. Hence, yes, you could (say) divide red and green by blue and come up with some other set of axes - which is close to what the standard chromaticity diagram does - but there is no "correct" choice - there is merely "convenient" choices. So you come swooping in and implying to our poor OP that everything he's already been told is wrong - for what reason? To make you seem 'smart'? Well, trust me - it doesn't. We commonly use RGB axes because that's closest to the biological truth (for daylight viewing conditions in the center of your field of view). We could get into MUCH more complexity by talking about vision at the periphery of your visual field - and in near-darkness and so forth - but we're trying to keep it simple and answer the question. So we're all answering the question using three perpendicular axes labelled "red", "green" and "blue" - and that's convenient and comprehensible. You are not making yourself sound at all "clever" by picking some wierd-assed coordinate system and then loudly proclaiming that everyone else is wrong. That doesn't fly - OK?
• Your explanation for the perception of violet is wrong. The high frequency 'bump' in the red curve is the reason.
• "Red green and blue are used for color reproduction" - not always. A color inkjet printer (for example) uses cyan, magenta, yellow and black for example. The color gamut of ink, CRT's, LED's, DLP's, color photographic emulsions, fabric dyes, etc are all different - but that's irrelevent because we're asking what humans can see - not what we're able to display in various media.

So: Your post doesn't advance answering the question in any meaningful way - it simply introduces confusion just for the sake of it...which I find annoyingly typical of several of your recent posts. SteveBaker (talk) 03:15, 19 February 2009 (UTC)

I think everyone involved in this discussion understands all of that. Whether you like it or not, the three cones are frequently referred to as "red", "green" and "blue". Even the paper I linked to above uses those names. It makes sense, too - red light stimulates primarily the red cones, blue light stimulates primarily the blue cones and green light is, well, somewhere inbetween. Those colours don't correspond to the peaks in the response curves, but that doesn't means they aren't useful names. Violet is between red and blue when viewed in terms of polar coordinates centred at the white point in xy-space because of the response curves. It's those response curves which determine the xy-space (as I explained above, the xy-space is defined in terms of colour matching functions that were determined using human response curves - those functions are the ones shown in this plot and you can clearly see a bump in the red curve). If you had different response curves, the curve corresponding to monochromatic light would be somewhere else, in a different shape and violet could be in a completely different place relative to blue and red. I don't know if it is necessarily the red bump that arranges the curve like that (the red bump in the paper I linked to is far less pronounced than the one in the CIE1931 colour matching functions, and there is also a lesser bump in green which isn't shown in CIE1931 - I'm not sure exactly what they did with the response curves to generate the colour matching functions), but it is to do with the response curves. --Tango (talk) 02:04, 19 February 2009 (UTC)

Hooray, I've offended everybody. This is depressing. I shouldn't have written in that for-dummies way, it was insulting. Please take it as having been meant for someone else, someone who really is a complete beginner in this. I'll try to be less annoying in the future.

But... I still can't see where you're coming from with a lot of this stuff.

• The XYZ coordinates don't approximate the LMS cone responses and weren't intended to—that wasn't one of the design criteria. (The design criteria were all coordinates nonnegative for real colors, efficient use of the space of nonnegative coordinates, and Y approximating luminance.) ${\overline {x}}$ has two peaks because it's a linear combination of ${\overline {\ell }}$ and ${\overline {s}}$. This doesn't help answer the question of whether ${\overline {\ell }}$ has a second peak. There exist linear combinations of the XYZ CMFs that are everywhere positive and don't have that peak, for example the combination given by the matrix at LMS Color Space#RLAB, so the XYZ CMFs are consistent with there being no secondary peak.
• "Violet is between red and blue when viewed in terms of polar coordinates centred at the white point in xy-space because of the response curves"—sure, but that's a very broad statement. Practically any set of three cone types will lead to a U-shaped curve, and white will always be somewhere in the middle of the disc formed by the convex closure since it's a combination of all the colors, so you will always be able to define a color wheel around white, and in any such color wheel the highest perceivable frequencies will appear to be intermediate in hue between the lowest frequencies and the not-quite-highest frequencies. There's almost nothing you can conclude about the cone responses from this wrapping-around behavior. The only thing substantially different that could happen would be if the curve doubled back in such a way that parts of it weren't on the convex boundary. I think that would be sufficient to prove the existence of a secondary peak in one of the cones. But the XYZ CMFs don't wrap around that way. Anyway, the perceptual effect of that would be different—it would mean something like high frequency light looking blue, higher frequency light looking violet, and even higher frequency light looking blue again.
• "Green is the color seen by the green sensor...by definition." I know some people call it the green cone, but I don't think even those people would say that it sees green by definition. Green means perceptual green or light with a wavelength around 550nm. If it were possible to stimulate the M cone alone, the visual appearance would be similar to supersaturated 500nm light, i.e. cyan, if the hues extend in the way one would expect. An LMS color system would be kind of a violet-cyan-magenta color system, with the violet right at the end of the spectral locus, the magenta at around (0.84,0.16), and the cyan literally off the chart (no xy coordinates because X+Y+Z < 0). It's fine as a color space but I have a hard time believing anyone would refer to it as RGB (does anyone have a source that does?).
• "The high frequency 'bump' in the red curve is the reason [for the perception of violet]." You're going to have to explain your reasoning, because I don't get it. If there's an extra bump in the L response then suppressing it would presumably change the curvature of the violet end of the boundary in xy coordinates. Maybe it would be straighter without the bump. But violet would still be violet. It would be a little more saturated but still pretty much the same hue—unless the end was shorter in which case it would be bluer, or longer in which case it would be purpler, but changing the curvature doesn't matter much. Let's do something more concrete and zero out the second peak in ${\overline {x}}$, even though that's much larger than any extra peak in L would be. That causes the cyan-blue-violet side of the boundary to become vertical, approaching (0,0) instead of (0.17,0) in the revised coordinates. Sky blue is also going to move toward the left, but the line from white through sky blue is still going to hit the boundary somewhere short of the very highest frequencies. So the frequencies higher than that will still have some red in them, assuming that blue is primary because of the sky. For that matter, look at the other edge of the spectrum. Psychological primary red isn't all the way at the end, so you might expect even lower frequencies to look a little bit like magenta, and in fact they do, at least in plausibly clamped spectra like this. That's definitely not due to a second peak in the S cone—the response of the S cone at those frequencies is indistinguishable from zero.
• There are four psychological primary hues, red, yellow, green, and blue. Stimulation of one cone type in isolation doesn't produce any of these hues—M alone gives you something between green and blue and S and L alone give you something between blue and red, and none of those hues looks primary. So I shouldn't have said that red, green and blue are irrelevant to color perception, since they're three-fourths of that RGBY system, but they're not any more primary than the other fourth of that system and they shouldn't be confused with either the cone outputs or the various systems of 3D RGB coordinates. Of course RGBY and sRGB and LMS are all fine as color systems. I'm not saying that any color system is good or bad, I just want to use distinct names for them since they are all different. In particular I don't want to use the name RGB for the LMS system, because it demonstrably confuses people. (Trovatore at least was confused by it, and Steve, I'm still not completely convinced you aren't confused by it also.) "Red, green and blue cones" is not quite as bad since it's unambiguous, but it's still very confusing to beginners (I think). A lot of people think that the cones directly measure coordinates in RGB space, and I want to make it as clear as possible that that's not true. Not because it's going to matter in most people's lives, but because telling people things they don't need to know is what the Ref Desk is all about. If the original poster did need to know the answer to this question, we probably would have deleted it since it would be a request for medical advice.

Anyway, friends? I hope? I still think you're wrong, but it's nothing personal... -- BenRG (talk) 21:54, 19 February 2009 (UTC)

Did you take a look at the paper I linked to? You can clearly see an increase in the responses of the L and M cones towards shorter wavelengths. I don't know the significance of them, but they're clearly there. You say pretty much any cone response curves will yield a U-shape, can you explain that? What's stopping there being a W-shape or even a straight line (which would make Steve's lecture title accurate - purple really would be a shade of green!)? --Tango (talk) 23:57, 19 February 2009 (UTC)

Actually, having given it some more thought, I can partially answer my own question. You would only get a straight line if the three cones were linearly dependant, which would be a very inefficient way of making an eye. --Tango (talk) 00:10, 20 February 2009 (UTC)

I think you can only get a W shape if at least one CMF has two big bumps in it. If they broadly respond to "high", "medium" and "low" wavelengths then you'll get a U. Even if it's a W the convex closure still looks like a "disc" and white is somewhere in the middle.

I just looked at the 1980 paper you linked. They're measuring absorbance of the photopigments, which is not quite the same as the cone fundamentals. I think the difference is that the cone fundamentals also take into account absorption by the lens and the macula. The photopigment spectra from here, which are based on papers from 1999 and 2000 (according to the site), also show an upswing, but it's at around 420nm, which is already well into the violet. The 1980 paper's results seem to show the upswing at around 450nm. I don't know the reason for that difference. The cone fundamentals from the same site (based on the same papers) show no upswing, since most of that light never reaches the cones anyway. It's not clear to me what should and shouldn't be interpreted as a peak, when it comes right down to it. I'd rather the peak not exist since that would show without a doubt that the argument for its existence is incorrect, but I still don't believe the argument even if the peak is there... -- BenRG (talk) 22:10, 20 February 2009 (UTC)

A W-shape would give a disc with white in the middle, but the hues would be in a different order. I think you would get red-yellow-blue-green-red (sticking with the same primaries as you were using above), with violet a bluey-green. (I'm working this out by visualising the colour space and kinda guessing - I should probably get to grips with the maths a little better and work it out properly...) --Tango (talk) 23:02, 20 February 2009 (UTC)

CIE 1931 color space chromaticity diagram equivalent tending towards black

The color question above made me think that there should be a triangle - analogous to the one above - with black at the center. Is there such a thing and what is it called. The two together should come closer to covering all colors visible to humans. I didn't want to stick this on to the above question because the two are only related. 76.97.245.5 (talk) 00:13, 17 February 2009 (UTC)

Only in a subtractive color model do colors converge on black (which is not used for light). So I doubt you'll find a colorspace diagram for it, as they are modeled on light, not paint... --98.217.14.211 (talk) 00:43, 17 February 2009 (UTC)

Provided you'd consider no light as black wouldn't different intensities of colored light give different results. The triangle seems to be all with the same intentity. (Oops there's two. Don't know in a hurry which one I should link.) 76.97.245.5 (talk) 01:19, 17 February 2009 (UTC)

The chromaticity diagram covers only 'chrominance' (ie 'color') and doesn't include 'intensity' or 'brightness'. To show all of the colors that humans can see you need a three-dimensional diagram with red, green and blue as the three axes. I have one of these on my desk at work - it's a perfect demonstration of 3D color space. You could certainly produce other diagrams as 2D slices through that cube at different angles. SteveBaker (talk) 01:28, 17 February 2009 (UTC)

The OP may also find Munsell color system (and some of the articles linked therein) of interest. Deor (talk) 12:53, 17 February 2009 (UTC)

Matter and Anti-Matter creation from energy

When matter is created from energy, is the outcome randomly determined (of it being matter or anti-matter)? And if so, would this explain the predominance of matter over anti-matter in this universe? 70.171.16.131 (talk) 06:19, 17 February 2009 (UTC)OP

As far as I know, which isn't much compared to some refdeskers, energy can ONLY be converted to matter if the corresponding antiparticles are created at the same time. You can't pick one or the other; that would violate conservation of charge, etc. (Please correct where necessary.) -- Aeluwas (talk) 09:41, 17 February 2009 (UTC)

When matter is created from energy, such as in the photon-photon collisions or from matter particles colliding at high speed at CERN and elsewhere, conservation laws specify that some quantities stay the same before and after the collision. Importantly, the number of particles minus the number of antiparticles must stay constant, so if 2 particles collide, you could for instance get 8 particles and 6 antiparticles out of that collision. Known violations of these laws are heavily studied but so far noone has been able to fully explain the predominance of matter over antimatter in the universe, it's one of the big questions in physics today. If you want to learn more, CPT symmetry or books on quantum mechanics may help. EverGreg (talk) 11:32, 17 February 2009 (UTC)

protein to build muscle mass

Hello Wikipedia,

In order to comply with current gay body-fascist requirements, i'm looking to build muscle-mass so would like to know what i should eat. The supplements people say about 1g of protein per pound of body weight but others suggest that this is cobblers and i should just eat sufficient calories so that i'm rarely hungry. My problem is that these people have an agenda to either sell me supplements or magazines and i don't know who i should believe. Does anyone know any objective figures? (ideally based on good ol' fashioned peer-reviewed science?)

Many thanks, 81.140.37.58 (talk) 11:16, 17 February 2009 (UTC)

The gays have created a body-fascist movement? Or the body-fascist movement itself is inherently homosexual? Or perhaps it's just disdained? Maybe the fascists just like gay bodies? This is all very confusing...

Regardless, "objective" advice (of which there is effectively none -- I doubt that peer-reviewed science is in agreement on this one) is likely to stray into medical advice. If you don't trust the subjective advice of magazines, I suggest you try the subjective advice of a doctor, physical therapist, and/or other health professional who can take your personal circumstances into account. — Lomn 14:19, 17 February 2009 (UTC)

The gay mafia must be behind the gay fascists. :-) But seriously, you need to exercise to build muscles. This will make you eat more and possibly crave protein (if you aren't already getting enough). You don't have to take any supplements or force yourself to eat more of any item to gain muscle mass, just eating what you want to eat will be sufficient. The exercise is the key. Eating more or taking protein/carb supplements without exercise will only make you fat. Unless you're going for the bear look, that probably won't help. StuRat (talk) 14:37, 17 February 2009 (UTC)

Do you go to a gym? If so, you can probably book a few sessions with a personal trainer and they will be able to give you nutrition advice (or, perhaps, refer you to a nutritionist) - they will be able to tailor it to your current diet, build, weight and your planed training routine and desired results. We can't do that. --Tango (talk) 14:45, 17 February 2009 (UTC)

So there's no general rule? Ho Hum... —Preceding unsigned comment added by 81.140.37.58 (talk) 17:09, 17 February 2009 (UTC)

When I researched this a few years ago, I didn't find much if any reliable scientific evidence in support of protein supplementation or even high protein diets. There seemed to be a minimal amount of scientific evidence in support of creatine supplementation. A Quest For Knowledge (talk) 17:10, 17 February 2009 (UTC)

Also, you might want to check out our Bodybuilding supplement article. A Quest For Knowledge (talk) 17:15, 17 February 2009 (UTC)

Ah, see I read an article like this and wonder why we can't theorize about medical treatment. I would think there is at least as much legal trouble in store by recommending creatine. I realize you didn't really recommend it and this isn't directed so much at the former user as it is at maintenance of a general principle. As far as this forum goes, please talk to a doctor and not a GNC representative about supplements for building muscle. Seriously, a MD. You wouldn't believe what they have found in weight gain supplements. About your other question, what you should eat, I would say you definitely need to run a caloric excess to build muscle. That simply means you eat more calories than you burn in a day. Remember that working out and running burn a lot of calories. Here is a website that automatically calculates your basal metabolic rate(BMR) using a popular equation. On top of your BMR you have to add any calories lost through exercise to get a good estimate. A protein rich diet is a common paradigm of building muscle as well. Basically the goal is to daily store protein as muscle. Now, this won't make you look that good, but it will make you strong. Professional body builders usually rotate evry 3-4 months between a caloric excess to a caloric deficit. The result in the long run is loss of fat and large well defined muscles.

--Mrdeath5493 (talk) 20:50, 18 February 2009 (UTC)

Most complex formula in Physics

What is the most complex formula that you know in Physics?--Mr.K. (talk) 11:55, 17 February 2009 (UTC)

The complex ones are boring, it's the simple ones that are interesting. You can get as complicated a formula as you like by just coming up with a really complicated problem to solve. --Tango (talk) 13:49, 17 February 2009 (UTC)

Yeah, you can make arbitrarily complex equations for whatever you want. However, you asked which is the most complex equation in physics that I know, so I'll tell you the most complex equation that I remember actually using. It is as follows:

$\rho {\frac {\partial ^{2}{\vec {\zeta }}}{\partial t^{2}}}=-\nabla (-\gamma p\nabla \cdot {\vec {\zeta }}-{\vec {\zeta }}\cdot \nabla p)-{\vec {B}}\times (\nabla \times (\nabla \times ({\vec {\zeta }}\times {\vec {B}})))+(\nabla \times {\vec {B}})\times (\nabla \times ({\vec {\zeta }}\times {\vec {B}}))+(\nabla \Phi )\nabla \cdot (\rho {\vec {\zeta }})$

(reference) which is the linearized magnetohydrodynamic equation of motion for plasma in terms of the plasma displacement vector, ${\vec {\zeta }}$. Of course, there are always more complex equations, but that's probably the longest one I've ever used. --Bmk (talk) 15:17, 17 February 2009 (UTC)

Magnetohydrodynamics will rank high on many people's lists because it has a lot of terms, effectively building up from "F=ma" for a single particle, accounting for all the forces involved. Plasmas encompass the many different domains of electromagnetism and Newtonian statistical mechanics, encompassing half of the fundamental forces recognized by the Standard Model. In some extra-special space-plasma cases, there can even be relativistic magnetohydrodynamics, adding another few terms. Other disciplines, such as subatomic physics, often introduce new conceptual ideas rather than adding additional terms to an equation. "Complexity" of a formula is sort of tough to decide; for example, if you wanted to solve the simple ballistic trajectories for thousands of catapults, the individual equations would be trivial but the implementation details to solve the equations simultaneously could be quite complex. At a certain point, physical equations cease to be represented as closed-form expressions, and will probably be represented as a computational physics simulation, numerical solver, or computer program. The complexity of such a representation can range in to tens of thousands of lines of computer code for numerical processing, data management, and user interface. Nimur (talk) 16:16, 17 February 2009 (UTC)

As I'm not a physicist, I don't have much experience of long formulae, but this one is pretty nasty (Continuous Fourier transform#Spherical harmonics)):

$F_{0}(r)=2\pi i^{-k}r^{-(n+2k-2)/2}\int _{0}^{\infty }f_{0}(s)J_{(n+2k-2)/2}(2\pi rs)s^{(n+2k)/2}\,ds.$

The Laplace transformations and Laplace's equation articles will give you ammunition to impress the ladies at parties. --Mark PEA (talk) 21:40, 17 February 2009 (UTC)

It's been a while since I used it, but $R=\kappa T$ (see Einstein's field equations) still gives me nightmares. Don't think it's complicated? It's actually (if I remember correctly) 4-dimensional, 4th order non-linear differential equation, with the only saving grace being a couple of symmetries. Confusing Manifestation(Say hi!) 22:52, 17 February 2009 (UTC)

Hmm, how about the Standard Model Lagrangian? -- BenRG (talk) 23:35, 17 February 2009 (UTC)

I don't know anything about the subject myself, but I've heard that workers in quantum chromodynamics routinely use computers to crunch formulas with millions of terms. They are sort of like Feynman diagram calculations in QED, except in QED the higher order terms become negligible quickly enough to make pencil and paper calculation tractable, but in QCD they need those huge computer calculations to get any actual numbers out. 207.241.239.70 (talk) 07:55, 18 February 2009 (UTC)

Microsoft Office Live Workplace for Linux users

This question has been moved to the Computing Desk, Nimur (talk) 16:31, 17 February 2009 (UTC)

Liquid density

How can this problem be solved: A mass of a liquid of density ρ is thoroughly mixed with an equal mass of another liquid of density 2ρ. No change of the total volume occurs. What is the density of the liquid mixture?

A.4/3 ρ      B.5/3 ρ
C.3/2 ρ      D.3ρ  —Preceding unsigned comment added by 116.71.33.96 (talk) 13:12, 17 February 2009 (UTC) 

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. SteveBaker (talk) 16:53, 17 February 2009 (UTC)

The Reference Desk will not do your homework for you. That said, I'll offer some pointers: Consider the volume of the first liquid to be x. What is the mass of the first liquid (there is a density-volume-mass relationship)? What is the mass of the second liquid? What is the mass of the mixed liquid? Now solve for density. — Lomn 14:14, 17 February 2009 (UTC)

Event horizon and neutron star crushing question

A recent Scientific American article about the possible existence of naked singularities got me wondering about neutron stars and event horizons. Specifically, there is a certain amount of gravitational force across the spherical boundary of an object needed to overcome the nuclear forces that keep a neutron star from collapsing into a singularity. There is also a certain amount of gravitational force to bring a sphere's surface area escape velocity to the speed of light, turning the sphere's boundary into an event horizon.

What's not clear to me is that those two levels of gravitational force are equal. So my question, then, is how much gravitational force is needed to start the process of crushing a neutron star into a singularity, and how much gravitational force is needed to create an event horizon around the surface of a collapsing object? And if those two numbers aren't equal, could it be possible to have an object that is, at least temporarily, significantly denser than a neutron star but whose escape velocity is less than the speed of light and thus is not a black hole? (A naked singularity is presumably one such object.) 63.95.36.13 (talk) 15:19, 17 February 2009 (UTC)

You may find Quark star interesting, but it's not entirely relevant to your question. For a object of a given mass you have something called the Schwarzschild radius. This depends only on mass, so for the neutron star/black hole in your scenario, it is constant. Once the neutron star collapses to smaller than its Schwarzschild radius, it is a black hole and has an event horizon, before then it isn't and doesn't. The gravity of a black hole isn't any stronger than the gravity of anyone else of the same mass, the only difference is that you can get closer it its centre of gravity. Therefore, it is meaningless to ask how much gravity is needed to create an event horizon, the correct question is how small does an object of a given mass need to be to form one. If the gravity is great enough to overcome degeneracy pressure and collapse the star, then sooner or later it will get small enough to form an event horizon and become a black hole. Its density will continue to increase until it passes that point (after which it becomes rather meaningless to talk about density). --Tango (talk) 15:30, 17 February 2009 (UTC)

Thanks for the tip about Quark Stars; that's the first I've heard of them, and it's actually related to my question in that it sort of shows an example of a star that is in an intermediate phase matter (at least at its core) between that of a neutron star and a singularity. As far as whether all singularities are black holes, though, that's precisely what the naked singularity article in SciAm was discussing (ie scenarios in which a singularity for one reason or another is not within an event horizon). Interesting stuff. 63.95.36.13 (talk) 18:30, 17 February 2009 (UTC)

Incidentally, have you read our articles: Naked singularity and Cosmic censorship hypothesis? (Lots of tags at the top of them both, so they may not be very good - I haven't read them recently.) --Tango (talk) 18:41, 17 February 2009 (UTC)

I did, but unfortunately they didn't really deal with the question above (or if they do it was in an over-my-head technical area). 63.95.36.13 (talk) 22:50, 17 February 2009 (UTC)

Tango: The gravity of a black hole isn't any stronger than the gravity of anyone else of the same mass, the only difference is that you can get closer it its centre of gravity — not strictly true, it matters where the mass is: shell theorem. —Tamfang (talk) 07:49, 20 February 2009 (UTC)

Coincidentally, we covered a similar question in january EverGreg (talk) 18:16, 17 February 2009 (UTC)

Nuclear submarine crash

What could actually happen if two nuclear-powered and nuclear-armed submarines crashed into each other way beneath the sea? On the one hand we have PR flaks saying, oh, nothing to worry about, the tea sloshed out of the mugs and we'll have to re-paint, that's all. At the other extreme we have ... nuclear war, possibly? (Worst case scenario: If one of the bombs were accidentally activated, and then another nation felt called on to respond.) Or the in-between possibility, of the vehicle being totalled, shades of Texas road crashes. BrainyBabe (talk) 16:40, 17 February 2009 (UTC)

Hmm. Are you asking because you've read this article (or another one on the same subject) or is this just a great coincidence, considering that was published yesterday? -- Aeluwas (talk) 17:00, 17 February 2009 (UTC)

Yes, I was in fact trying to tie together X and Y, but I guess absolute explictness trumps collegial allusiveness. BrainyBabe (talk) 17:12, 17 February 2009 (UTC)

I'm just surprised there was not a need for a major rescue. When Kursk (submarine) hit a rock (or blew a dud torpedo, as the current consensus seems to hold), everyone perished despite best efforts to send help. To me the most amazing thing about such a scenario is the total information-vacuum. The capacity for a submerged submarine to communicate with the surface is limited (exact data rates are probably classified), but I suspect we are talking about a few hundred bits per second as an absolute maximum. This is barely enough to even send a distress call, let alone a full explanation of the scenario. And to imagine that these guys must have a special bit-sequence to initiate second strike! Let's all hope their bit-error rate is REALLY low. Between the highly-classified nature of submarines and simple logistics which limit the number of witnesses, I think the truth of any submarine accident is always pretty elusive. Nimur (talk) 17:20, 17 February 2009 (UTC)

The chance of a nuke detonating accidentally is close enough to zero as makes no odds. Smashing them, shooting them, blowing them up, whatever, won't have any effect, you have to fire neutrons into a critical mass of fissile material - that just doesn't happen unless someone makes it happen. The real risks (to the world - obviously there are risks to the submariners) are leakage of radioactive material (pretty unlikely) and loss of nukes which could then be recovered by terrorists or rogue nations (everything is so top-secret that they would have to have spies in just the right places to find out where to look, so this is a pretty low risk too). There is also a risk that one side may not believe that it was an accident and retaliate, but they wouldn't retaliate with nukes (loss of a submarine does not warrant mutually assured destruction), so you're just looking at a regular war, at worst. --Tango (talk) 17:55, 17 February 2009 (UTC)

Delving away from the scientific aspects of this question into the political aspects, it should be noted that France and UK are allies and have been for quite some time. It is unlikely a war would erupt between these two nations even if a nuke were detonated. Had this occurred between the US and the Russia, the situation might be quite different. I suspect a war would not break out, but there would be accusations and counter-accusations. Of course, if this had happened during the middle of Cuban Missile Crisis, all bets are off. A Quest For Knowledge (talk) 18:18, 17 February 2009 (UTC)

It was only after the collision and reports started coming in that it was worked out what had happened - neither knew who or what they had crashed into at first. If a nuke had gone off, there would never have been any reports and the UK and France would each have simply seen a nuclear detonation at the approximate location of one of their nuclear subs. What conclusions they would have drawn from that, I don't know, but they might well have acted before letting the other know they had a nuclear sub in the area, so before anyone could work out what had happened. --Tango (talk) 18:28, 17 February 2009 (UTC)

You may be amused/terrified to learn that deliberate provocations along these lines *did* occur during the Cuban Missile Crisis, including forcing Soviet missile subs to surface, and actually test-firing an ICBM! That's what you get with the Buck Turgidsons of the world running the show. --Sean 21:59, 17 February 2009 (UTC)

OK, so far so good, but I posted on the science desk deliberately. What happens when a sub hits something (another sub, or a mountain for that matter), at various velocities? What would it take to knock a hole in one? What sorts of damage could an undersea collision cause? Could it become "stuck" down there, without massive other damage? BrainyBabe (talk) 19:05, 17 February 2009 (UTC)

I heard on the BBC Radio 4 this morning some submarine walla talking about a submarine that several years ago crashed into an undersea pinnacle at 30 knots. A crew member was killed and several others received serious injuries but the shell of the submarine remained intact and there was no damage to the nuclear fuel cell on board. Funny that we didn't hear about it. Richard Avery (talk) 19:50, 17 February 2009 (UTC)

You're referring to USS San Francisco (SSN-711)#Collision with Seamount? --Carnildo (talk) 02:18, 18 February 2009 (UTC)

Failure modalities:

• the sea-hull is damaged, and/or control surfaces are damaged and function improperly: the sub must limp to its home port; if the damage is bad, it must surface and call for a tender
• the prop is damaged, the driveshaft or its bushings bent or damaged, or the gearbox damaged - mild gearbox damage may be reparable or patchable at sea, for everything else the sub cannot make way and must call for a tender (it can remain submerged while it waits)
• the sea hull is deformed to an extent that the hatches to the pressure hull (man hatches, torpedo tube hatches, missile tube hatches) leak, seals for the prop shaft leak, and/or the ballast system piping and its valves are damaged and leak - pumps can handle small leaks (if the sub remains at shallow depth); for more serious leaks the affected compartments must be evacuated and sealed, and the sub must surface
• large scale damage to the pressure hull, or major damage to the ballast system - compartments flood and must be quickly sealed off to prevent the sub sinking. With more than a few areas waterlogged, or if the balast system is badly damaged or inoperable, the sub cannot surface. Crew must be rescued (the US Navy uses a DSRV, I don't know what arrangements if any the British and French navies have). 12-36 hours pass before rescue begins, and as the DSRV has a capacity of ~8 people, rescue of the boat's 100+ compliment takes some time. If the electrical systems have failed the air is not reprocessed and becomes rich in exhaled CO2; the crew resort to breathing apparatus with wearable scrubbers. If these become exhausted, or if insufficient are available (particularly if many crew are trapped in a small compartment) crewmembers begin to die from CO2 poisoning (it's my understanding that you die from CO2 poisoning before the hypoxia kills you).
• damage to the pressure hull affects many compartments; too much of the boat floods for the ballast system to lift, and the boat sinks to the bottom of the ocean. In sufficiently deep water the remaining pressure hull fails under the pressure and the crew is killed.

Submarines are always very sturdily constructed to withstand the pressures they must face; military submarines (which must also withstand violent maneuvers and near-misses from depth charges) particularly so. Naturally they don't publish crash-test results, so we can't say which of the above scenarios are likely to result from collisions of a given speed. As noted above, it is very unlikely that a nuclear warhead will explode. There is a larger chance that the solid rocket motors of an SLBM will explode, or that the reactor encapsulation would be torn open to the sea, but it's difficult to imagine a non-explosive impact that could trigger either of those without also shattering the pressure hull and killing the crew anyway.

87.115.43.168 (talk) 20:19, 17 February 2009 (UTC)

The film Gray Lady Down gives a fictional account of the rescue of the crew from a striken US attack submarine; despite being a "disaster movie", it's probably rather optimistic. Some of the crew of Kursk were alive for days after she was damaged; it would appear the Soviet->Russian Navy did not have an effective rescue technology (a DSRV sounds great on paper, but if the sub's hatches are damaged, or if she is lying at a funny angle, or if she's taking on water faster than the rescue equipment can arrive, then things are more complex). 87.115.43.168 (talk) 20:28, 17 February 2009 (UTC)

Wow, 87.115, you sure know your stuff! I'd buy you a drink if I could. BrainyBabe (talk) 21:17, 17 February 2009 (UTC)

Resolved

Per documentaries, it is the responsibility of the sub personnel to monitor the sound generated by other vessels to know what sort of subs or surface vessels are near them. A nuke sub is basically pretty loud and distinctive. Even without using active sonar, which would give away your position, you should know when another sub is on a collision course, and emit a sonar blip, or use underwater audio communications to warn off a friendly nation's sub or take evasive action so as not to hit an unfriendly nation's sub. Some skippers' careers are likely to have passed their zeniths. A collision of subs is apt to be a disaster, and smacks of negligence, like a collision of airplanes. Edison (talk) 05:58, 18 February 2009 (UTC)

Aren't military subs designed to be as quiet as possible? --Tango (talk) 14:47, 18 February 2009 (UTC)

French Defense Minister Herve Morin claimed " "These submarines are undetectable, they make less noise than a shrimp." His claim seems highly suspect. A large group of shrimp all snapping is actually quite noisy. A shrimp can produce a reported 218 db noise. "As quiet as possible" still leaves them with the Russian Sierra class at a http://forum.nationmaster.com/forums/viewtopic.php?f=7&t=740 reported] 120 decibels and the U.S. Los Angeles class at 110 decibels. By comparison, a loud rock concert is reported to be about 115 decibels, a power saw at 3 feet 110 db, and a motorcycle 100 db. There is a lot of heavy equipment operating on a submarine while it is in motion. Another source on the noise levels of various nations' subs, both diesel, battery and nuke, is at [1]. Why can't a sonarman or equivalent detect the equivalent of an approaching noise source of this magnitude? Edison (talk) 20:02, 18 February 2009 (UTC)

I guess there could be some chance of damaged reactors making a radioactive mess. As for bombs going off, the permissive action links should prevent that. (Update: Um, except on looking at that article, urk, the Brits apparently don't use them. This is technology that the US deliberately disclosed to the USSR in the 1960's to decrease the chance of accidental war, so I thought everyone used them.) 207.241.239.70 (talk) 08:01, 18 February 2009 (UTC)

The road to Xibalba

The lead section of Xibalba states: Another physical incarnation of the road to Xibalba ... is the dark rift which is visible in the Milky Way. Is this an example of a dark cloud constellation? Astronaut (talk) 18:03, 17 February 2009 (UTC)

The dark rift is nothing more than gas and dust obscuring our view of the lens-shaped galaxy. Because we're on the outer edge of the galaxy - we're seeing all of this 'stuff' edge-on and we're looking through a lot of it. Just as humans have often discerned patterns in stars ('constellations' - the 'signs of the zodiac', etc) - we also see shapes in the clouds (it's ALWAYS a "bunny"!), in the pattern of marias on the moon, faces on Mars - and so it's no surprise that people have seen (and named) patterns in these gigantic dust clouds. It's an example of pareidolia. SteveBaker (talk) 22:35, 17 February 2009 (UTC)

I was thinking more specifically. Xibalba is a Mayan mythology, and the Dark cloud constellation article mentions the Incas, whose mythology appears to contain some concepts common to the Mayans. I was considering adding mention of the Mayans to the Dark cloud constellation article. Astronaut (talk) 09:54, 18 February 2009 (UTC)

Well, if you are planning on changing one of our articles - you'll need some suitable references that state this fact explicitly. IMHO, yes, this is a reasonable example of that...but that's not good enough - you need a solid reference. SteveBaker (talk) 13:59, 18 February 2009 (UTC)

I just want to know, exactly where is the bunny in the picture above? ~Amatulić (talk) 21:42, 18 February 2009 (UTC)

Health: green tea leaves v. cabbage leaves

I've been drinking green tea recently. In the teapot, after a while the dry tealeaves swell up and can be clearly seen to be parts of a leaf. I'm wondering if there is anything special about the tea plant: would infusing or even eating the same amount of eg cabbage leaves be as good for you as green tea is supposed to be? 78.151.117.148 (talk) 19:07, 17 February 2009 (UTC)

I can't directly answer your question, but I went to school with our local swimming champion, who drank a litre of cabbage water every lunchtime: the school dinner ladies saved it for him! When I asked him why, he told me that the vitamins from the cabbage were actually in the water rather than in the cooked cabbage, and it was better for you than eating the cooked cabbage!--TammyMoet (talk) 21:25, 17 February 2009 (UTC)

Potential effects of tea on health makes many claims for green tea's health benefits; Cabbage makes fewer. They are quite different plants and they presumably contain many different trace chemicals. Tempshill (talk) 21:27, 17 February 2009 (UTC)

Green tea provides ECGC, a powerful antioxidant. ECGC may protect against diabetes. Green tea has a reputation for increasing metabolism and reducing the risk of obesity. Cabbage is a cruciferous vegetable; cruciferous vegetables such as broccoli and cabbage are associated with a lower risk of cancer. Green tea and cabbage might both be considered superfoods.151.213.161.26 (talk) 23:37, 20 February 2009 (UTC)

Gold in volcano

I recall reading about an active volcano in Antarctica that has a large amount of gold in the volcanic ash. Is this just a tall tale, or does such a volcano exist? If so, which volcano is it? 65.167.146.130 (talk) 20:04, 17 February 2009 (UTC)

"Large amount" is rather overstating the case, and our article on Mount Erebus doesn't mention it, but see this New Scientist article (which is the first hit in a Google search for Antarctica volcano gold). Deor (talk) 20:16, 17 February 2009 (UTC)

sort of alchemy

Ok, the ancient alchemists dreamed about turning lead into gold. I have heard that this is now possible using a particle accelerator or something and bombarding lead to break off protons and neutrons. Is the gold created in this manner a stable isotope, or will it be radioactive? 65.167.146.130 (talk) 20:13, 17 February 2009 (UTC)

I believe you are referring to nuclear transmutation. Yes, it is possible but is too expensive to be worth the cost. A Quest For Knowledge (talk) 20:35, 17 February 2009 (UTC)

And if it were inexpensive to do, the supply of gold would increase substantially, which would lower the price of it. --Mark PEA (talk) 21:28, 17 February 2009 (UTC)

yes, but is the resulting material radioactive, or stable? —Preceding unsigned comment added by 65.167.146.130 (talk) 21:34, 17 February 2009 (UTC)

Per our article on the synthesis of noble metals, Au₁₉₇, the only stable isotope of gold, can be produced from mercury. Likely there exists a path from lead to Au₁₉₇, but it's also quite likely more complicated. — Lomn 22:01, 17 February 2009 (UTC)

Why is there a high concentration of catalase in the liver?

What specific purpose does the liver serve by having a high concentration of the enzyme catalase? Does the body pump most of the waste hydrogen peroxide produced to the liver to be treated there? Or is there another reason?

22:01, 17 February 2009 (UTC) —Preceding unsigned comment added by Scofield Boy (talk • contribs)

I don't know for certain, but it may play a role in detoxifying potentially harmful compounds and/or removing peroxides from the blood. – ClockworkSoul 22:34, 17 February 2009 (UTC)

I can't find a definitive answer, but I believe, essentially, what Clockwork says. Peroxisomes are responsible (in whole or in part) for some functions that take place mainly in the liver (gluconeogenesis and fatty acid metabolism). I suspect this means the liver needs more peroxisomes, and also more catalase (many of the reactions involved in those processes are oxidative, and so could create free radicals and superoxides that would also lead to peroxide). Someguy1221 (talk) 23:14, 17 February 2009 (UTC)