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December 18

Hydrochloric acid and methylbenzene

Why is hydrochloric acid acidic in everything except methylbenzene? I am really confused about this as my textbooks do not include anything of the sort and I have been told that this would appear in my exam. Please help. Thanks! pcfan500 (talk) 06:28, 18 December 2014 (UTC)[reply]

You'll probably want to review the concepts about what makes things "acidic" - see Brønsted–Lowry acid–base theory and Lewis acids and bases for an overview of the two big ones. In short, a compounds is an acid because it donates a proton (accepts an electron pair). Hydrochloric acid is a strong acid because the chloride doesn't hold on to the proton very well ... at least in water. Other solvents (like the aforementioned methylbenzene) don't do as good of job stabilizing the separated state, so the proton and the chloride want to stick together. I'm not sure why methylbenzene is being highlighted specifically - I'd expect other similar solvents (ethylbenzene, for example) to also behave similarly, sot it's not really the case that hydrochloric acid is acidic in "everything" but methylbenzene. -- 141.39.226.228 (talk) 08:45, 18 December 2014 (UTC)[reply]

why is time known as the fourth dimension?

why is time known as the fourth dimension? 94.98.4.75 (talk) 09:37, 18 December 2014 (UTC)[reply]

I've added a heading to this question. It's known as the fourth dimension because it is one in a physical sense, i.e. subject to the rules of physics in a very similar sense. Of course you can also see the sense in which it!s a dimension: just consider cases where we remove a dimension and use it for time instead, like a flipbook of animation . There are two dimensions shown at a time and the third dimension gets shown over time. 91.120.14.30 (talk) 11:23, 18 December 2014 (UTC)[reply]

To fully describe where an object can be found, you need 4 dimensions, with the 4th being time. Consider trying to describe the location of something which no longer exists, like the Lighthouse of Alexandria. You can go there now, but you won't find much. StuRat (talk) 11:37, 18 December 2014 (UTC)[reply]

You might (or might not) be interested in the technicalities given in articles such as Spacetime and Minkowski space. Dbfirs 12:37, 18 December 2014 (UTC)[reply]

(edit conflict) Geometry "works" with any numbers of dimensions. You know how there are all sorts of "math" you can do with shapes and lines and such? Calculating area, length, volumes, velocity, etc.? Well, the "rules" that allow you to make those calculations are not restricted to any number of dimensions. For example, you can establish a line in two dimensions by defining two sets of points, say point A = [0,0] and point B = [1,2]. You can then set up equations in either cartesian coordinates or vector coordinates to define the line that goes through those points. Well, you can do the same in 3 dimensions by defining the points in 3D space as A = [0,0,0] and B = [1,2,3] or some such, and then can write an equation to define that line. Now, the deal is, even though you can't picture a line in any more than three dimensions, the rules for writing the equation of a line still apply in any number of dimensions. You just do the algorithms and define the line. I can define a line in 4 dimensional space simply by saying A = [0,0,0,0] and B = [1,2,3,4], and the rules for writing the equation for THAT line (which has no reasonable PICTURE, but never mind that) are the same rules as writing lines in less dimensions. I can have any arbitrary number of dimensions, and the math for describing an object called a "line" in those number of dimensions is the same as it is in 2D, 3D, or whatever. So that brings us back to why even bother to treat time as a dimension like space dimensions: that is, we have the three spacial dimensions (up-down, left-right, forward-backward) and add time as a fourth number into that set. The reason why has to do with Einstein's theory of special relativity. What special relativity shows (among other things) is that you can vary how you move through time. Just as you can move through space at various rates, it turns out that time passes at different rates depending on certain conditions, such as the mass and velocity of an object relative to nearby objects (the effect of mass on time passage is actually covered by general relativity, not special, but whatever). Now, it turns out that because the rate of passage of time for an object is variable just as it's movement through space is variable, in order to completely describe the motion of an object, one needs to consider not only how it's position is changing, but also how it's timescale is changing with respect to other objects. In order to do that, you treat time like a dimension, and do your calculations in 4D rather than 3D; but the rules for doing so (as noted above) still apply. One last thing about time, however, is that the time "dimension" doesn't behave like the other "dimensions": it has it's own set of rules which is different than the others; however as long as you take those rules into account, you can still do math with it to predict the behavior of objects (and that's what physics is: the science of being able to predict the behavior of objects in motion). The specific set of dimensions (which includes the three spatial dimensions and the one time dimension) we use to do these calculations is called Minkowski space, named after the mathematician who worked out the math of such a system. The last question someone might ask is why do we have to do all that. The answer is because it is necessary to explain observable phenomena where normal 3D "Cartesian/Euclidian" space cannot, for example experimentally verifiable phenomena like time dilation, or the invariance of the speed of light. I know this was a little TL;DR but I hope it is clear enough to help one understand the entire point of treating time like a dimension. --Jayron 32 12:39, 18 December 2014 (UTC)[reply]

Your "TL" discussion makes sense. The way I like to "picture" the time dimension is to think of the state of the universe (or some portion of it) at a series of points in time - as with the flip-book discussed earlier. So the fourth point in that [x,y,x,t] coordinate system can be pictured as what that [x,y,z] system looks like as "t" changes. Beyond that, of course, it gets tricky trying to picture. But in math, as you say, you can have any number of dimensions and the equations still work, albeit getting more and more complicated with added dimensions. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:52, 18 December 2014 (UTC)[reply]

Actually imaginary time is the fourth dimension. The distance in relativity calculations seems to be determined by x² + y² + z² - t². Wnt (talk) 14:34, 18 December 2014 (UTC)[reply]

Not really. Normal, everyday time is the fourth dimension. Imaginary time is used only for very specialized calculations, such as to eliminate the singularity (division by infinity) at the Big Bang or in black holes. --Jayron 32 15:17, 18 December 2014 (UTC)[reply]

I'd say that normal everyday time is a fourth dimension. It's definitely the most common choice in any non-technical context. As you discuss above, there are lots of other choices for what dimension we might call the fourth (or fifth, sixth etc.) especially if one is delving into theoretical physics (e.g. string theory) or certain mathematical structures (e.g. octonians)SemanticMantis (talk)

Yes, "the" in this case refers to "the fourth dimension used in Minkowski space for relativity purposes". Imaginary time as a fourth dimension only has limited utility in understanding a few physical phenomena. --Jayron 32 15:58, 18 December 2014 (UTC)[reply]

My understanding of the "spacetime interval" (see Spacetime) is that we can treat time as a fourth dimension for calculating a distance if we treat c as the conversion factor between our measurements, and recognize that when measuring time we are measuring multiples of i. Wnt (talk) 19:14, 18 December 2014 (UTC)[reply]

I think the consensus nowadays is that treating time as an imaginary spatial coordinate is a cute mathematical trick, but probably too cute, because it seems more meaningful than it is. I believe there's a note on it in Gravity by Misner, Thorne, and Wheeler, a book we should probably have an article on if we don't already. --Trovatore (talk) 21:05, 18 December 2014 (UTC)[reply]

Per Trovatore, it's a matter of perspective. Because of the sign conventions of working in Special Relativity, the "time" dimension has the opposite sign of the spatial dimensions. Mathematically, this means that some of the terms have a value of i attached to them. The math is identical if you attach the i to each of the three spatial dimensions, and leave time in the real number set; or if you attach i to the time dimension and leave the three spatial dimensions in the real numbers. Conventionally, we tend to leave the i in the time dimension because it makes the math a bit easier (in the sense that we have one imaginary number and three real numbers), but time itself is a real number. The use of imaginary time only comes in, if I am not mistaken, in unusual situations where the standard sign convention [-,+,+,+] for [t,x,y,z] produces physically paradoxical results (such as singularities). At least, that's my understanding. --Jayron 32 21:18, 18 December 2014 (UTC)[reply]

And we do have an article on the book. It's called Gravitation. Not Gravity. --Jayron 32 21:21, 18 December 2014 (UTC)[reply]

Well, what I'm thinking is that if you use a two or three dimensional coordinate system, you can say that within that coordinate system, every two points has a defined distance between them. That distance doesn't change unless you look at it from a frame where the whole coordinate system is changed. But in a system of "four dimensions" with real time, the distance between any two points is not constant, but depends on the frame of reference of whoever is looking at it. So to say time is the fourth dimension in that case is sort of meaningless. I mean, you can make the color of the object the fourth dimension, if you're willing to have a coordinate system that you can't calculate distance in. But use time * i as the fourth dimension and you do have a real distance that is Lorentz invariant. Wnt (talk) 21:25, 18 December 2014 (UTC)[reply]

Here's a less coordinate-dependent way of phrasing things. It requires you to know a little differential geometry. The point is that the metric tensor has three positive eigenvalues and one negative one. This is just a fact; it's not based on which arbitrary coordinate system you choose.

The x₄=ict trick is sort of an attempt to obscure this fact, or if not actually an attempt, risks obscuring it. --Trovatore (talk) 21:29, 18 December 2014 (UTC)[reply]

Alright, I'll admit it... I'm in the fog here. I'm afraid I'm missing how spacetime coordinates have four eigenvalues in the first place. Wnt (talk) 00:22, 19 December 2014 (UTC)[reply]

Because there are four dimensions to move in: up-down, left-right, forward-backward, and past-future. --Jayron 32 01:32, 19 December 2014 (UTC)[reply]

Wnt and Jayron, I think you're both confusing imaginary time (as mentioned, for example, in A Brief History of Time) with the old (now disfavored) convention of using a pure imaginary value for the t coordinate instead of a flipped sign in the metric. They are different. "Imaginary time" is kind of the opposite of what you're thinking: it starts with a mixed-sign metric with all coordinates including t real-valued, and then considers (unphysical) imaginary values of t to make the metric effectively Euclidean. -- BenRG (talk) 19:30, 19 December 2014 (UTC)[reply]

Changes in Mammalian Milk composition - Biochemistry question

I understand that a cattle's food, environment, and artificial hormone shots can change the composition of it's Breast Milk. Any professional name for this phenomenon? I need it to efficiently search for some literature in this subject. Thx. Ben-Natan (talk) 13:10, 18 December 2014 (UTC)[reply]

One source suggests "biochemical alterations" in breast milk after heating. I'd think the preferred term in that case is denaturation. Though, it's certainly not what you're looking for. Let me see if I can find a better term. 71.79.234.132 (talk) 14:46, 18 December 2014 (UTC)[reply]

I just searched google scholar for /diet nutrition effect cow milk/ - These articles were near the top of the list [1] [2] [3]. From skimming the abstracts, it does not seem that there is a single term to cover all the food/environment/hormone effects on cow's milk. The term "ruminal biohydrogenation" and "Conjugated linoleic acid" are used quite a bit, and the keywords used by the articles should help further searchers, e.g. "mammary metabolism", "fatty acid desaturation" "milk fatty acids." The first linked ref above should be especially useful, as it is an Annual Reviews article, which are usually an expert summary of a broad field of research and give lots of references. SemanticMantis (talk) 15:33, 18 December 2014 (UTC)[reply]

circuit analysis

How do I solve the first order differential equation for an LR circuit WITHOUT using Laplace transforms ie from first principles? — Preceding unsigned comment added by 109.152.195.34 (talk) 13:46, 18 December 2014 (UTC)[reply]

The canonical solution for a linear first-order ordinary differential equation is a solution using the separation of variables method. In the case of an L-R circuit, you'd have a first-order equation in current with respect to time, parameterized by the inductance and resistance.

When I write out every step, this procedure takes longer than simply applying the Laplace transform by inspection, so in practice, mathematically-inclined people tend to memorize the solution of a simple circuit (instead of explicitly re-solving it). You must simply recognize the standard form, understand the relationship between the relevant variables, and recall the standard-form solution.

Nimur (talk) 15:27, 18 December 2014 (UTC)[reply]

How do the enzymes and nutrients in breast milk survive body temperature?

This source suggests that heating breast milk can denature some enzymes and nutrients. I haven't read the full article yet, so I can't tell what temperature they set the breast milk at. But human body temperature is 37 degrees Celsius. Can't the breast milk's enzymes survive when exposed to some heat but not too much heat? Would it be better for women to take stored breast milk from the refrigerator and heat it up with their own body temperature? Or would they have to acquire a wet nurse? 71.79.234.132 (talk) 14:57, 18 December 2014 (UTC)[reply]

Self-evidently those proteins do not degrade at body temperature; they are made at body temperature, stored at body temperature, and consumed at body temperature. --Jayron 32 15:14, 18 December 2014 (UTC)[reply]

Obviously this can't be a problem. We (and all other mammals) have evolved to do this without any refrigeration or whatever. Also, any degradation due to a brief period at body temperature would happen in the baby's mouth, throat and stomach anyway. Clearly the problems with heating milk (any milk, actually) happens at much higher temperatures. Efforts to (for example) sterilize milk might well suffer from this problem. SteveBaker (talk) 15:44, 18 December 2014 (UTC)[reply]

The temperature required to denature the proteins in breast milk would be roughly the same as required to cook an egg. The change in going from raw egg to cooked egg is mostly the process of protein denaturation.

Different proteins denature at different temperatures. In fact, a common experimental technique to measure the stability of the protein is to look at the circular dichroism of a protein as a function of temperature. (See also Protein_folding#Circular_dichroism). For most proteins you see a sigmoidal transition from "folded" spectra to "unfolded" spectra, with a characteristic transition point at a defined temperature. This temperature is called the "melting temperature" of the protein, and varies from one protein to another. Some are very unstable, and will unfold at or around room temperature (mostly proteins from psychrophiles). Some are stable at room temperature (25 C) or body temperature (37 C) but will unfold at 45 C or so. Different proteins unfold at different points, all the way up to 95+ C, where it becomes hard to measure. (There are proteins which don't unfold even under boiling conditions - mostly these are from thermophiles, but there are some mutants of proteins from mesophiles which have very high melting temperatures.) - So the answer to the original question is that there's a large swath of temperatures between 37 C and 100 C, and there are some proteins which are stable at 37 C but which will unfold at 45 C or 55 C or 65 C or 75 C, etc. And the temperature at which the unfolding/denaturation happens for one protein is not indicative of what will happen for other proteins.-- 141.39.226.228 (talk) 10:40, 19 December 2014 (UTC)[reply]

Resistance of electrically conductive paint.

I've been looking at these electrically conductive paints:

  http://www.solianiemc.com/assets/Specifiche/Conductive-Paint-Specification.pdf

...and trying to find out how much resistance I'd get if I painted strips of varying widths.

It quotes the resistance in units of ohms/sq - I have no idea what 'sq' means...square meter? square millimeter? Then the values are 0,3 (which I suspect is 0.3 in one of those places in the world where they use '.' and ',' in the opposite sense to the more common US/UK useage).

Anyway, if I use a layer of the stuff of the recommended thickness to paint a 'wire' that's N mm wide and some much longer length - what kind of resistance would I measure per mm of length for various values of N? (This seems like it might be a variant of: http://xkcd.com/356/ ...in which case, I apologize in advance!)

TIA SteveBaker (talk) 15:35, 18 December 2014 (UTC)[reply]

“

The reason for the name "ohms per square" is that a square sheet with sheet resistance 10 ohm/square has an actual resistance of 10 ohm, regardless of the size of the square.

”

- from Sheet_resistance#Units I don't know how to compare the resistance of an Nx1cm strip to a Nx1m strip, but this seems to say rather clearly (if counter-intuitively) that the resistance of a NxN square is equal to the resistance of a (2N)x(2N) square. It's unclear to me if the resistance would be different for a (2N)x(2N) square compared to a (N)x(4N) rectangle (but I'd guess they would be different). If you figure it out let us know :) SemanticMantis (talk) 15:47, 18 December 2014 (UTC)[reply]

Oh...that's strongly counterintuitive! So a square that's a mile by a mile has the same resistance as a 1" x 1" square?! I guess that as the distance increases, so do the number of parallel paths that the electrons can travel though...so the two numbers cancel out and the resistance stays the same. Weird!

So if the resistance of an NxN square is always the same - then I could mentally chop my 1cm wide strip of paint into 1cm squares that are in series and say that an N cm long by 1cm wire has N times the resistance of a 1cm x 1cm strip...which is just the ohms/sq number?

Which would mean that the ohms per meter of a strip of this stuff is inversely proportional to the width...which seems entirely reasonable.

Resistance = ohms/sq * length / width ?

If someone could confirm my intuition on this one, we can call it "answered". (And thanks to User:SemanticMantis for a great & fast reply).

SteveBaker (talk) 16:40, 18 December 2014 (UTC)[reply]

Upon further reflection, the geometry is probably more important than the area. So a circle of any area will also have the same resistance, but it will be different than the square resistance. And once the proportions of a rectangle are fixed, they should have the same resistance independent of area as well, I think... SemanticMantis (talk) 18:54, 18 December 2014 (UTC)[reply]

Sounds right. Read the second paragraph of the section linked by SemanticMantis. It says you just multiply the square resistance by the aspect ratio to get the resistance for a rectangle. Proving the exact result for painted traces with corners or other bends would be tricky, but I suspect that total length over width is still a good approximation. 12.195.117.49 (talk) 19:25, 18 December 2014 (UTC)[reply]

Specifically, SteveBaker, resistance is proportional to the length and inversely proportional to the cross section. With paint, the cross section is, for all intensive purposes, the same as the width, and by definition for a square, the length and width are the same. μηδείς (talk) 01:33, 19 December 2014 (UTC)[reply]

Ah...yeah - that makes sense. ...BTW: the phrase is: "all intents and purposes"...not "all intensive purposes".

Thanks everyone...I think I have everything I need. SteveBaker (talk) 04:17, 19 December 2014 (UTC)[reply]

Cloning by chance

This is a topic I've read about before, but I recently came across it in a graphic novel, so I'm interested in recalling the specifics: Given the size of the human genome, what are the chances of an individual being born with DNA identical to that of another individual? The story in question posits an interplanetary population of 100+ trillion humans, and one of the characters claims that "three people are born with my DNA every day," which seems impossibly high. Like I said, I know I've read about this idea in scientific literature before; just not sure what keywords to use or where to start looking. Evan ^{(talk|contribs)} 18:57, 18 December 2014 (UTC)[reply]

The human genome is about 3 Gigabases. So, there is around 4 to the power 3,000,000,000 possible combinations. Not, all of them are, of course, actually possible but this rough estimate still holds. So, the claim that "three people are born with my DNA every day" is false if the population is around 100 trillion. Ruslik_Zero 20:06, 18 December 2014 (UTC)[reply]

That's theoretically true - but some of those possibilities would imply that the mother gives birth to a tree or a duck or a elephant. Totally random DNA isn't likely to arise in a population. An oft-quoted number is that 99.9% of my DNA is identical to yours (or to any other human) - so taking that rough number says that only 3,000,000 base pairs are really likely to vary between people. Still, that's 4^3,000,000 - let's say 10^1,000,000 to pick a nice round number. Given that there are only around 10⁸⁰ atoms in the visible universe - it's still SPECTACULARLY unlikely that two people would ever have the same exact DNA by chance reshuffling of A's, G's, C's and T's.

However, we have to consider that the man and woman (who love each other *very* much and make some babies) each only have 23 pairs of chromosomes - their child doesn't get a random selection of A's, G's, C's and T's from each parent - it gets entire chromosomes. So for any given pair of humans, there are only 2²³ possible chromosomally unique children that they can have...about 8 million. If those children were to in-breed, so no new chromosomes appear then their children would still only have some combination of their grandma & grandpa's DNA. So if our 100 trillion humans were all descended from Adam and Eve, there would only be 8 million unique human chromosome combinations - and there would indeed be a bunch of people with the same DNA. However, copying errors, mutations and the fact that it's been a hell of a long time since our most recent common ancestors guarantees that there is considerably more variation than that.

I don't buy the story's claim...but it's not so simple to dismiss it as that. SteveBaker (talk) 20:55, 18 December 2014 (UTC)[reply]

Note chromosomal crossover in meiosis is essentially a required event for proper gamete production (it can be omitted, but only with a significantly greater chance of abnormalities as I recall). This means that there are vastly. vastly more than 2²³ outcomes. Wnt (talk) 21:18, 18 December 2014 (UTC)[reply]

Depends on how you're defining "the same". A DNA database will try to claim an absurdly low probability, but will only look at as many markers as are needed to reach that -- and there is a risk that in a particular small ethnic group the different markers will not truly be independent, but will be more likely each to go a certain way. This risk seems to be very low, but it is not zero - consider the trivial case where a person turns out to be the identical twin of someone who was secretly swapped at birth, which however soap opera unlikely is not astronomically unlikely. However, the ways in which this can happen are fewer the more markers are examined. The identical twin will always come out the same, but fellow 100% Tasmanian aborigines will eventually be distinguished, assuming any known method of reproduction.

I still have in the back of my mind a nagging doubt whether it could ever happen that humans clone themselves naturally, if a diploid egg or sperm were to provide all the genetic material to the exclusion of the other gamete. Such embryos normally die, but that observation only holds up until a counterexample can be found. But with the number of large-scale genetic tests on the general population this is rapidly fading from absurdly unlikely to genuinely ruled out. Wnt (talk) 21:14, 18 December 2014 (UTC)[reply]

How much would it cost get tested for every genetic risk factor known to man?

How many are there (including the minor ones like propensity for male pattern baldness)? Hundreds? Thousands? Where would one go? Would that be an unusual request there? I guess they could give you the list of risky genes that can take effect before 30 (hundreds?). You'd take it home, cross out the ones that'd be unoverlookable by your age (bubble boy syndrome, complete immunity to chickenpox..) and only get tested for genes that could bother you before testing gets cheaper. How much would that cost? I don't think I'd actually do it though, I'd wait till it's cheaper and understanding of genetics isn't so piecemeal. Well, if it's tens of dollars (yeah right) I'd consider it but I'm curious how much I'd have to have to not mind spending the money. Also, how much is it to find out just the known genes for cancer? breast(40s) — colon, skinny male breast (!) is my parental history of cancer. I guess a single disease is cheaper to test for than many. (grandpa died from cigarette cancer at 56 before we could find out whether he would've died from regular cancer, other grandpa died a year after being well enough to make baby the original way, increasing the chances that it was cancer, but he might've smoked) Sagittarian Milky Way (talk) 22:23, 18 December 2014 (UTC)[reply]

That's not a question that is easy to answer, as there are many genetic risk factors about which we know, but which we don't know (i.e. we know they are inheritable, but we don't know which combinations of genes cause them). If you thrown in a research program to identify them all (or even just a few of them), the sky is the limit. On the other hand, if the genetic factors are well understood, testing all of them should not be impossibly expensive - a quality whole genome sequence cost Steve Jobs US$100000, and prices have come down significantly. I think the analysis of the genome should be highly automatable, at least in principle. --Stephan Schulz (talk) 22:55, 18 December 2014 (UTC)[reply]

Your question implies there is a test for every genetic risk factor known to man, but not every gene or set of genes that causes a disease that runs in a family has been identified. This is about the third time since halloween that we've had this question, you might want to serach the archives for lengthy previous answers. μηδείς (talk) 01:18, 19 December 2014 (UTC)[reply]
- Thanks Medeis, but what search terms should I use? I can't seem to find one, much less two after Halloween. Sagittarian Milky Way (talk) 15:00, 19 December 2014 (UTC)[reply]

You may want to look into 23andMe and in particular their pre November 22, 2013 test kits (probably available at elevated prices on eBay). Ariel. (talk) 08:13, 19 December 2014 (UTC)[reply]

It would probably cost you your life. When you add up the X-ray exposure, exposure to radioactive tracers, exposure to chemicals, tissue damage due to biopsies, blood drawn for hundreds of tests, etc., the net result is a pretty large insult to the body. Looie496 (talk) 15:37, 19 December 2014 (UTC)[reply]

He meant "genetic" and I have fixed the title to reflect that. μηδείς (talk) 19:24, 19 December 2014 (UTC)[reply]

- I think you're describing getting tested for every "disease with a non- nurture component known to man". Clearly no one is going to want a piece of their lung (much less every organ) pulled out of their body and many other invasive tests in their 20s with no evidence of malfunction. And I actually imagined "every blood test known to man, including thousands of toxins" as a child, pictured hundreds of vials of blood and thought the image amusing.

- If you could get a full genome without too much somatic expense then you should easily be able to get enough DNA for fewer genes. Sagittarian Milky Way (talk) 16:40, 19 December 2014 (UTC)[reply]

See $1,000 genome. Even so, be skeptical, because medicine is a racket. The 23andMe action corresponded with other efforts to have it declared "unethical" to get genetic testing done without getting certain common specific genetic tests done, starting with the infamous BRCA1. However, shortly after that decision the Supreme Court ruled against a broad class of gene patents, creating some hope again. Even so.... I think that one way or another, someone will step forward to demand a huge amount of protection money before the average person is allowed to find out about his genome, because that's how medicine works. Wnt (talk) 21:46, 19 December 2014 (UTC)[reply]

In your hair

After seeing a shampoo advert, I was wondering, what does caffeine do to your hair? Would the same results occur if you used tea, coffee or a high caffeine drink e.g. Monster or Red Bull? What would be the effects of each of those? Also why do some people pour beer or another alcoholic drink in their hair? 5.69.204.149 (talk) 23:34, 18 December 2014 (UTC)[reply]

Beer is supposed to give hair body. You can google beer shampoo for that answer. What caffeine does is wash out and run down the drain. μηδείς (talk) 01:23, 19 December 2014 (UTC)[reply]

Its just bogus marketing. There is no benifit in caffeine on your hair. They just play men to belive "activating" will prevent the loss of hair at advanced age but since this is a natural process caused by hormones "activation" will likely even speed up that natural process. --Kharon (talk) 11:58, 19 December 2014 (UTC)[reply]

Caffeine may not do much to your hair, but it can be absorbed through the skin [4]. There are products like caffeinated soap [5] that purport to deliver the drug through the skin. So if there's a lot of caffeine in shampoo, absorption through the scalp may deliver similar effects to taking caffeine orally. SemanticMantis (talk) 18:02, 19 December 2014 (UTC)[reply]

Anything that would deliver a clinically measurable dose of caffeine through the skin would have a lethal dose of caffeine per mouthful. Given how often people consume soap and shampoo, there'd be a flurry of deaths and an episode of Inside Edition with Bill O'Reilly exposing the danger. μηδείς (talk) 19:22, 19 December 2014 (UTC)[reply]