Wikipedia:Reference desk/Archives/Science/2010 October 25

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October 25

Dark spots/bruises on old people's skin

What are the dark spots and bruises that old people (60+ years old)? Most of them look like dried blood under the skin or a bruise. What are they? Why are they there? 76.169.33.234 (talk)Dave —Preceding undated comment added 03:10, 25 October 2010 (UTC).

Age spots? DMacks (talk) 03:15, 25 October 2010 (UTC)

I looked at pictures and I don't think thats it. They are purple or black spots, some are small, some are large. I've seen them mostly on old people's arms, but I think they many be all over, not sure. —Preceding unsigned comment added by 76.169.33.234 (talk) 04:48, 25 October 2010 (UTC)

Aging skin is susceptible to a number of conditions that present themselves as dark spots. The lentiginous conditions are are a range of pigment depositions which may be benign, pre-malignant or malignant. Older skin may also be more susceptible to injury and bruising, for a number of reasons, including medication effects, bleeding and clotting disorders, and age-related thinning of the epidermis and dermis. I would encourage anyone with an undiagnosed pigmented or dark spot on their skin to seek a doctor's opinion in real life, not on wikipedia. Mattopaedia ^{Say G'Day!} 05:53, 25 October 2010 (UTC)

Many elderly people take blood thinners or anticoagulants such as Plavix (to counter the risk of stroke and heart attack) which can result in much easier and sometimes quite severe bruising. -- 117.47.204.34 (talk) 11:57, 25 October 2010 (UTC)

Yes, as I said, this is one of a number of possible causes. Again, see a doctor if you don't know what the cause or nature of the "dark spot" is. Not all the causes are innocuous. Mattopaedia ^{Say G'Day!} 14:26, 25 October 2010 (UTC)

Thanks for the answers. I don't have it, I just see it on a lot of my older customers at the pharmacy I work at. I didn't get a chance to ask the pharmacist so I thought I would ask here. 76.169.33.234 (talk)Dave —Preceding undated comment added 05:05, 26 October 2010 (UTC).

Calorimetry

6.60g of $CaCl_{2}.6H_{2}O$ were dissolved in 100 mL of water. The temperature dropped 0.895°C. What is the molare heat of dissolving? ${\frac {mc\Delta T}{n}}={\frac {100*4.184*0.895}{6.60/219.1}}=12419Jmol^{-}1$. This is not accurate because a lot of energy goes into the vessel, the glass rod/thermometer etc. If the 100 mL of water in the calorimeter fell by 0.700°C when 4.00A was passed through at a potential difference of 3.00V for 35 s.

I've been told the calibration factor is ${\frac {VIt}{\Delta T}}$, which yields ${\frac {3*4*35}{0.7}}$ = 600 J°C^-1. How do I use this to find the molar heat of dissolving? Do I just multiply this by ${\frac {\Delta T}{n}}$ so $600*{\frac {0.895}{6.60/219.1}}$?--MrMahn (talk) 10:09, 25 October 2010 (UTC)

Atom pic

I need a pic of an atom that may be used to illustrate that much of the volume of anything consisting of atoms is actually empty since the nucleus is small (compared to the atom) and the electron cloud is very sparse. I looked at the atom article, but the illustrations there aren't very helpful. -- kainaw™ 15:57, 25 October 2010 (UTC)

In ballpark figures, the atomic radius is around 10,000 times larger than the nuclear radius. So in any realistically scaled diagram that scales the whole atom to the width of your screen, the nucleus will still be too small to see. Maybe some animation zooming in from atom to nucleus would illustrate their relative sizes better than a static diagram. Gandalf61 (talk) 16:14, 25 October 2010 (UTC)

Powers of Ten is one such animation. -- BenRG (talk) 21:10, 25 October 2010 (UTC)

An atom isn't mostly empty space. Okay, electrons are pointlike in a certain sense. But quarks and gluons are pointlike in the same sense, so in that sense an atom is entirely empty space, as is everything else. To the extent that there is such a thing as occupying space, atoms do it; they're the quintessential example of something that does. It's true that most of the mass is concentrated in a small volume, but there's no threshold mass density at which space becomes full. -- BenRG (talk) 21:10, 25 October 2010 (UTC)

I agree. The atom isn't mostly empty space. What is true is that an atom has a very small collisional cross-section, as compared to the atomic spacing. Probably, if you want to be accurate, you want a plot of scattering-angle probabilities from a Rutherford scattering experiment (surprisingly, we do not have this plot in our articles, but I can look one up in a textbook and create it later tonight if it's necessary). Or, if you prefer to use a classical approach, consider one of the diagrams in the Bohr model article. "If drawn to scale", the atomic "solar-system" diagram will look like you expect - mostly sparse - but it isn't exactly a correct picture of the atom. Nimur (talk) 21:25, 25 October 2010 (UTC)

The point-like nature of the electron is an abstraction. I mean, if you want to measure "how big the electron is", you get something based on its wavelength, which is on the scale of an atom. There's no way to measure how big the "point-like spot" is. When a white dwarf compacts a star into an object the size of the Earth, it's a million times denser than water, because it's made up of electron-degenerate matter where electrons no longer form neat little atoms in contact with one another but are all smooshed together, but it's still nowhere near as dense as a neutron star. So even there the electrons aren't "points", and they can't be made to get out of the equation without actually joining up with protons to form neutron-degenerate matter.

I'd say, as an opinion, that the most sensible way to think of the size of an atom is just to take a piece of ordinary matter, divide by the number of atoms, and there you are. There really aren't many spots of vacuum at the atomic level - molecules of water, air, etc. work their way into any significant crevice, and things crystallize in a way to make the atoms happy. They're elastic - they'll expand a bit if you atomize them (so to speak) into space, and they'll compress depending on the amount of force - but they're not empty space, except in the physics-philosophical sense that everything is empty space when you ignore the fields and phenomena that you use to describe the existence of matter. Wnt (talk) 22:25, 25 October 2010 (UTC)

Dry CuCl

How can I isolate powdered solid pure copper(I) chloride? I would like to take a picture for the article here. Thanks, --Chemicalinterest (talk) 16:02, 25 October 2010 (UTC)

That may be tough to do, as I would suspect (but not guarantee) that under normal conditions, the copper (I) chloride would disproportionate into metallic copper and copper (II) chloride, or possibly oxidizde in air just to make copper (II) chloride. That's why the only pic we have in our article is the green-tinted powder. As it is dried, its oxidizing away. It may be possible to do it in a completely inert (nitrogen or argon) atmosphere, assuming it doesn't disproportionate. --Jayron32 00:50, 26 October 2010 (UTC)

The picture on the right claims to be copper(I) chloride. It's a different green tint from the blue-green copper(II) chloride, but I can't vouch of its authenticity. Maybe a picture of a flame-test like the ones on [1] could be an alternative way to represent the compounds? EverGreg (talk) 08:57, 26 October 2010 (UTC)

Flame test? The copper(I) will oxidize in the flame. --Chemicalinterest (talk) 14:29, 26 October 2010 (UTC)

It's two or three years now that we removed that image from the copper(I) chloride article after complaints. I remember one professor saying that, if he received CuCl that looked like that from a chemical company, he would send it straight back! The problem is that the sample in the image is so oxidized as to be pretty much useless. Pure CuCl is white; the stuff you find in the bottle at the back of the shelf in the storeroom is often grey (because of oxidation). It's tough to avoid oxidation when you're preparing it if you don't have access to an inert atmosphere: I'd say it's more hassle than it's worth (although I'm sure Chemicalinterest wants to try!) because it's a fairly common lab chemical, and someone will take a photo at some stage. Physchim62 (talk) 09:50, 26 October 2010 (UTC)

I don't have access to an inert atmosphere though. --Chemicalinterest (talk) 14:29, 26 October 2010 (UTC)

Would heating it with ascorbic acid do anything? --Chemicalinterest (talk) 14:31, 26 October 2010 (UTC)

Most likely not. The copper(I) oxide is most likely the only substance which is easily to obtain.--Stone (talk) 18:25, 26 October 2010 (UTC)

When do you think they'll come out with plasma weapons?

Wondering... —Preceding unsigned comment added by 166.109.0.155 (talk) 16:54, 25 October 2010 (UTC)

This is a common question; see here, here and for a related discussion here. You might also be interested in the particle beam article. -- Finlay McWalter ☻ Talk 16:57, 25 October 2010 (UTC)

The main problem is that when people hear "plasma weapon" they think "Star Wars, cool." But in reality they would have pretty limited application, and as crude as it is, the plain old bullet (accelerating a piece of metal to high speeds) works better in most situations than a plasma (or laser) would. There are a few situations where the possibly high accuracy and speed of a laser weapon would be better (e.g. shooting down a missile) but in most cases, the tried-and-true method seems pretty clearly superior. --Mr.98 (talk) 17:37, 25 October 2010 (UTC)

The reference desk does not answer requests for opinions or predictions about future events. See information at the top of this Project Page. Dolphin (t) 00:52, 26 October 2010 (UTC)

Most likely they would not be as useful as you think on earth because the atmosphere will quickly kill the power of your plasma. If you had a plasma gun in space, perhaps that would be more useful. As far as practicality goes, I don't think you will ever see widespread plasma weapons on earth, though some research group might make one just to see if it can be done. Googlemeister (talk) 15:31, 26 October 2010 (UTC)

2 errors ?

1) In http://en.wikipedia.org/wiki/Doppler_effect#Analysis

How is the third formula developed from the previous two ? Isn't a square root missing in the third formula ?

2) In http://en.wikipedia.org/wiki/John_Strutt,_3rd_Baron_Rayleigh#Bibliography

The first two English links appear to be reserved for PC users. Netscape 7.02 over Mac OS 9.2.2 crashes after Google's initial screen blinks by.

http://www.archive.org/stream/theorysound02raylgoog/theorysound02raylgoog_djvu.txt (New York university) gives correct access to volume 1 for Mac users.

83.226.97.246 (talk) 17:06, 25 October 2010 (UTC)

1.) No, there shouldn't be an square root. 67.78.137.62 (talk) 17:13, 25 October 2010 (UTC)

(1) I don't believe the third formula is supposed to be derived from the previous two. But the previous two can each be derived from the third formula, by setting the appropriate variable to zero. So at least they're consistent.

(2) I didn't have any trouble with them on my Linux machine running Firefox. My bet is that any user of Netscape 7 is going to have those same problems. Netscape Navigator 7.02 is about seven years old, and Firefox (its spiritual successor) has gotten a lot more stable over the years. I haven't seen a page that crashes Firefox in a long time. Unfortunately, it looks like Firefox isn't available for OS 9. I searched around a bit and found iCab and Classilla (which I'd never heard of before); I bet one of those would work a lot better for a lot of pages, because support for browsers as old as that is declining rapidly among web developers. Paul (Stansifer) 17:29, 25 October 2010 (UTC)

Deriving the third equation from the first two is quite straightforward. It's useful to define additional variables, to keep $f$ and $f_{0}$ from being overused. Define $f_{s}$ as the frequency of the wave as of a moving source, $f_{r}$ as the frequency of the wave as of a moving receiver, and $f_{m}$ as the frequency of the wave as of a point in between the source and the receiver, which is stationary with respect to the wave's medium. The first formula in the article section gives the frequency as of a stationary point given the frequency as of a moving source, which using the variables we defined is

$f_{m}=\left({\frac {v}{v+v_{s}}}\right)f_{s}$

The second formula in the article section gives the frequency of a wave as of a moving receiver given the frequency as of a point that's stationary with respect to the wave's medium, which using the variables we defined is

$f_{r}=\left({\frac {v+v_{r}}{v}}\right)f_{m}$

Substituting the expression for $f_{m}$ given by the first equation above into the second equation above gives

$f_{r}=\left({\frac {v+v_{r}}{v}}\right)\left({\frac {v}{v+v_{s}}}\right)f_{s}=\left({\frac {v+v_{r}}{v+v_{s}}}\right)f_{s}$ ,

which is the third formula in the article section. Red Act (talk) 19:09, 25 October 2010 (UTC)

P.S. If you're seeming to remember that the equation for the ratio of frequencies in the Doppler effect involves a square root, perhaps that's because you're remembering the equation for the ratio of frequencies in the relativistic Doppler effect, which does have a square root. Red Act (talk) 19:19, 25 October 2010 (UTC)

Response: 1) Thanks very much for the clarificatons. However I feel even more, that a rewrite of the original article is needed. Your 'adjustments' of the denotations are more than adornments. They correct misleading denotation, especially in light of the last line of the section Analysis:"... but hold for the relativistic Doppler effect as well. Furthermore the use of the expressions "the wave" and "stationary with respect to the wave's medium" is obscure. 2) I do now not believe my troubles at obtaining a text from the links I mentioned originate with my using Netscape 7.02 for Mac . Why would I otherwise succeed in acquiring the text I required, when I simply used an alternative parallel source from the same http://www.archive.org/search.php?query=theory%20of%20sound ? It has only been in cases of programming deficiencies, that Netscape 7.02 for Mac has failed me. —Preceding unsigned comment added by 83.226.97.246 (talk) 23:20, 25 October 2010 (UTC)

Sticking the Landing

If a person were strapped inside the man-sized hollow chamber of a large ball of foam rubber and dropped from high enough to reach terminal velocity upon impact with the ground, how large would the ball of foam rubber need to be to not injure the person inside? TheFutureAwaits (talk) 18:19, 25 October 2010 (UTC)

You're asking how much foam rubber padding would be required for a person to survive a 200 mph impact unharmed? Considering the amount of crumpling metal required to protect someone in an auto accident at 65 mph, my first instinct is to say "lots", but I'm sure someone can come up with a more sophisticate response. the real problem (honestly) is more a matter of thickness then absorbency - a person traveling at 200 mph can only decelerate so quickly without suffering catastrophic injuries, and the rate of deceleration is primarily dependent on the distance over which deceleration occurs - I'm too lazy to do the math, but try calculating the amount of force needed to change velocity of a 160 lb person by 200 mph over the space of (say) one meter, then ask yourself if that would leave you feeling happy afterward. --Ludwigs2 19:10, 25 October 2010 (UTC)

Don't forget he said "terminal velocity". If you make the ball large enough it'll be a parachute of sorts. You can make the ball large, with thin rubber. Or smaller with thicker rubber. There are many many kinds of foam, each with different levels of impact protection. It's not really possible to calculate this, at best maybe someone knows a terminal velocity formula for a ball. Ariel. (talk) 19:34, 25 October 2010 (UTC)

I just assumed that a ball was relatively aerodynamic, so it would have an equivalent terminal velocity to a plummeting human. I could obviously be wrong. plus, you could fill the ball with helium to cut down terminal velocity further, though you'd talk funny when you got out. --Ludwigs2 20:01, 25 October 2010 (UTC)

Yah, but even so the ball can be made much larger than a human, but still weight about the same. Ariel. (talk) 20:32, 25 October 2010 (UTC)

There is also a great dependency on the orientation of the person at the time of impact. If the impact occurs with the person's head lowest and his feet highest the ball would have to be of very large diameter indeed to protect the person's head and spine. If the impact occurs with the person's feet lowest and his head highest that would be a little better than the previous case. The best configuration would be if the ball had a parachute or tail that ensured it remained orientated so the person was always lying flat looking upwards - the human body can withstand much greater acceleration or deceleration in that position than in any other. During the launch of astronauts and cosmonauts, when very high accelerations are employed for a long period of time, the astronauts and cosmonauts are lying on their backs. Dolphin (t) 21:37, 25 October 2010 (UTC)

Same general principle as lying (not standing) on a bed of nails. WikiDao ☯ (talk) 23:16, 25 October 2010 (UTC)

We need to decide the maximum acceleration we can let the person inside the ball be exposed to in order to answer the question. 2g? 10g? Physchim62 (talk) 02:42, 26 October 2010 (UTC)

Assuming the foam compresses nice and evenly to bring the test subject from an initial speed v to rest over a distance d at a constant deceleration ng, then

$d={\frac {v^{2}}{2ng}}$

Taking v as 200 mph or approx 90 ms^-1 and n as 2, I get d to be about 200 metres. Taking n to be 10, this reduces to 40 metres. Still quite a large air bag ! Gandalf61 (talk) 13:07, 26 October 2010 (UTC)

You might even be able to get away with 15 or 20g's if the human is in the back down landing configuration, but obviously the risk would be even higher. that might get you as low as 20 meters. Googlemeister (talk) 15:28, 26 October 2010 (UTC)

how would we even know

if we found a dinosaur fossils that jesus had wridden on, how would we even know? —Preceding unsigned comment added by 84.153.201.252 (talk) 18:35, 25 October 2010 (UTC)

sorry for making a alert! I just want to know, syentificly how would we know if we found a dinosor fossil like that! —Preceding unsigned comment added by 84.153.201.252 (talk) 19:05, 25 October 2010 (UTC)

We wouldn't. there's no way to tell from fossil remains the particular activities of the animal (except in very general ways). if you personally rode a horse, could you tell that you had ridden it from examining its bones? You might be able to determine if the horse had been ridden extensively or left to run in the fields (I assume that over time riding may make some detectable changes in the bone structure of the horse), but there would be no way to determine which or how many people had ever ridden the horse. --Ludwigs2 19:14, 25 October 2010 (UTC)

It also depends on your definition of Jesus. Depending on your particular views, Jesus was a specific individual who lived approximately 1 A.D. until approximately 33 A.D. (though even these dates are considered dubious, and most scholars (scientific and theological) believe Jesus was born about 4 years "before Christ." (Rather, it is the Gregorian calendar that incorrectly calculates the years since Jesus' birth). If you subscribe to this point of view, you would be able to use radiocarbon dating to determine the age of the dinosaur bones: if the fossil age does not overlap with the period of time that Jesus was alive, there is no way that he could have ridden on the animal. Most scientists will say with a resounding degree of confidence that radiocarbon dating is a well-established and accurate method for dating fossils that are around 2,000 years old; they will note a clear lack of any species of dinosaur that was alive at that time; and will conclude that it was impossible for Jesus to ride a dinosaur. (Indeed, most dinosaur bones are so old that their carbon signature is too weak to accurately use for dating; other forms of radiometric dating exist for older types of fossil, including dinosaur bones. If your views of theological Jesus are different from those of established scientific and historical evidence, you will have to explain them in a clear and empirical way before we can evaluate their scientific merits (if they have any). Nimur (talk) 20:12, 25 October 2010 (UTC)

(edit conflict)Considering that dinosaurs went extinct around 65 million years ago, and Jesus of Nazareth lived about 2000 years ago, there wouldn't be dinosaurs around for him to ride. Modern man arose 50,000 to 100,000 years ago, which is about 3 orders of magnitude after the dinosaurs; early man arose 2.5 million ears ago, 1 order of magnitude afterwards. CS Miller (talk) 20:18, 25 October 2010 (UTC)

I suppose it's not worth pointing out that even according to a very literal reading of the Bible, there should not be any Jesus-dinosaur overlap. --Mr.98 (talk) 20:32, 25 October 2010 (UTC)

Incorrect. Some flavors of Christianity, in particular the Orthodox sects, view Jesus as fully man and fully god who has always existed; in other words, "Jesus" is both son and father, in the Holy Trinity. Other flavors of Christian thought prefer to view Jesus as distinct (the "son" of God), and while of "one substance", not the same as God. Still other flavors of Christianity, particularly various Protestant factions, completely discount the notion of "trinity" and consider Jesus as totally distinct from God. See Religious perspectives on Jesus. It's futile to describe any of these as the literal interpretation of the Bible. They are all considered "literal interpretation" of the Bible. "The Bible" is self-contradictory, and contains passages to support all of these points of view. Furthermore, there is great debate about which translation is the authoritative version of the Bible; these quibbles usually boil down to exactly one of these types of interpretative schisms. A thorough survey of the different sects of Christianity will unveil that (in addition to political reasons), the various factions split from each other because of fights about these subtle distinctions. For example, a particular Syriac Orthodox Christian faction might very well contend that literal interpretation means that Jesus has always existed, and always will exist, and Jesus was one and the same with the Creator God of Genesis. Thus, Jesus always existed and could have ridden dinosaurs many millions of years before Christ. An American Baptist Christian would deny this wholeheartedly, and insist that literal interpretation of the Bible clearly indicates that Jesus is distinctly and only the son of the God Creator of Genesis, and did not exist until four years before Christ, when he was summoned into existence to die for the sins of mankind. A different American Methodist faction might contend that Jesus existed in spirit but was not on Earth until 4 before Christ. A Catholic would insist that Jesus has always existed but did not take the form of flesh until four years before Christ. All of these are well-founded in literal readings of Bibles of various languages and translations. Scientists do not believe that any humans co-existed with dinosaurs. Nimur (talk) 20:44, 25 October 2010 (UTC)

All birds are dinosaurs and therefore are dinosaurs not extinct so your calculation are clearly wrong. Our dinosaur article states "Consequently, in modern classification systems, birds are considered a type of dinosaur — the only group of which that survives until the present day." I do not think Jesus have ridden any birds since they are generally to small. --Gr8xoz (talk) 20:42, 25 October 2010 (UTC)

Ostriches are commonly raced in some countries, but they are native to sub-Saharan Africa. Somehow, I doubt this is what our queriant had in mind. CS Miller (talk) 20:52, 25 October 2010 (UTC)

(EC) 'generally' but not always - [2] [3]. Also during Jesus's time the Moa was still around. Of course Jesus wasn't near anywhere ostriches are found AFAWK, let alone moas who AFAWK weren't known to any humans at the time. Nil Einne (talk) 20:56, 25 October 2010 (UTC)

And let us not forget the elephant bird of Madagascar which I think is the size of a young tyrannosaurus rex and could be ridden if it didn't kill you first.

With only the information provided to go on, it would have to be by faith alone. WikiDao ☯ (talk) 00:39, 30 October 2010 (UTC)

Probability amplitudes

Why does $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$? I understand that it has to do with making $\langle \psi |\psi \rangle =1$, which is certainly convenient, but it can't all be just convention. $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$ implies that the probability for a state φ to be in the state ψ is equal to the probability for a state ψ to be in the state φ, which seems very non-trivial. How can this assumption be justified? 76.68.247.201 (talk) 20:49, 25 October 2010 (UTC)

I believe that equivalence is only true if the operators are Hermitian adjoint. As such, it is a matter of definition, not an assumption. See our article on invertible matrix, specifically:

• For any invertible n×n matrices A and B $\left(\mathbf {AB} \right)^{-1}=\mathbf {B} ^{-1}\mathbf {A} ^{-1}$.

The notation is not the same; but the derivation is similar. Nimur (talk) 21:19, 25 October 2010 (UTC)

Sorry, but what do operators have to do with it? Aren't these states? 76.68.247.201 (talk) 21:53, 25 October 2010 (UTC)

The rules of Hermitian conjugation apply equally well to states as to operators. I should say, I have never been a fan of Dirac notation, because this ambiguity does not help simplify the problem (it just makes it use fewer characters to write). Nimur (talk) 22:15, 25 October 2010 (UTC)

I think there's something I'm still missing. If you say that a bra is the hermitian conjugate of a ket, then isn't that the same as saying $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$? Certainly you can't just define the bra to be the hermitian conjugate of a ket, no? 76.68.247.201 (talk) 23:05, 25 October 2010 (UTC)

Yeah, there aren't really any operators directly involved in this question. Quantum mechanical states are vectors in a Hilbert space. The Hermitian operators used on that Hilbert space have a different dimensionality from the vectors in that space, just like matrices have a different dimensionality from vectors in ordinary linear algebra in Rⁿ. Red Act (talk) 23:29, 25 October 2010 (UTC)

I think this is the quantum mechanical version of the time-reversibility of physical laws, and as such it's reasonable to call it a physical assumption. It's justified in the same way as any other assumption, by consistency with experiment. That's not to say that you could drop it and get a quantum mechanics without time reversibility. It's too fundamental to the whole mathematical structure of the theory. -- BenRG (talk) 22:47, 25 October 2010 (UTC)

Sorry, how does $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$ imply time reversibility? And I thought quantum mechanics wasn't time reversible, what with the weak interaction and all. 76.68.247.201 (talk) 23:05, 25 October 2010 (UTC)

You define $\psi \rangle ^{*}=\langle \psi$ because it is Hermitian adjoint. Then, you have, by distributive property:

$\langle \phi |\psi \rangle ^{*}=\langle \phi ^{*}|\psi ^{*}\rangle =\langle \psi |\phi \rangle$

I think that does it, no? I have a more elaborate "rigorous" scheme, that involves pre-multiplying and post-multiplying and canceling out a bunch of terms that exactly equal 1, but that's a lot of TeX to type for essentially the same result as the above. Nimur (talk) 23:13, 25 October 2010 (UTC)

I think things have been defined differently for me than they have for you, so I'm having some trouble following your argument. For me, the "dagger" operation † can be defined such that |Ψ>^† = <Ψ|. I imagine my "dagger" is your *, but then you apply * to the amplitude <φ|ψ>, in which case * should be meant as the complex conjugate...so my guess is that * applied to an amplitude means something different than * applied to a ket. But then you say that <φ|ψ>* = (<φ|)^†(|ψ>)^†...isn't that tantamount to saying <φ|ψ>* = <ψ|φ>? Unless I'm missing something obvious (and I have a horrible feeling I am), then what you said seems circular. 76.68.247.201 (talk) 23:47, 25 October 2010 (UTC)

You can also think of the ket vectors being the elements of a Hilbert space and the bra vectors as linear functionals that map elements from the Hilbert space to the set of complex numbers. Then the set of all these functionals form a Hilbert space too (the so-called dual of the original Hilbert space). You then want to define a mapping between the Hilbert space and its dual. We can choose this mapping such that any ket vector |psi> is always mapped to that bra vector <V| which, when applied to |psi>, maps it to its squared norm. It is convenient to denote that particular bra vector <V| that |psi> is mapped to by <psi|. It is easy to see that this mapping from the Hilbert space to its dual can be chosen to be anti-linear.

If you now write down the expression for the squared norm of |a> + |b>, you find the desired result. Count Iblis (talk) 00:24, 26 October 2010 (UTC)

That works IF we force <ψ|ψ> to be real (and it seems that <ψ|ψ> = 1, as noted above), which you're obviously doing by making it the norm-squared of |ψ>. My question is, why is it justified that <ψ|ψ> = 1 or real or whatever? It has to be more than definition, because setting <ψ|ψ> = 1 has a physical significance beyond the mathematical formalism, namely that the probability for a state φ to be in the state ψ is equal to the probability for a state ψ to be in the state φ. 76.68.247.201 (talk) 01:08, 26 October 2010 (UTC)

That these inner products give the probabilities is postulated by QM. We postulate that the Hilbert space structure applies to Nature. There is no proof that this has to be true. You can prove the Born rule postulate, i.e. that |<a|b>|^2 gives the probability that a system in state |a> is found in state |b>, from weaker assumptions, see e.g. Zurek's proof. Count Iblis (talk) 02:01, 26 October 2010 (UTC)

But the only requirement is that |<a|b>|² is real. But what's stopping <a|a> from being complex? You could argue that, by definition of an inner product, <a|a> must be a real number greater than zero, but that just shifts our burden because now we have to show that every property of the inner product matches with observation. I'm okay with saying that |<a|b>|² = probability, because that's just the definition of the amplitude. I'm okay with saying that $\langle \psi |\phi \rangle =\sum \langle \psi |i\rangle \langle i|\phi \rangle$ because that follows from 1) the adding of amplitudes when an observation isn't made (justified by the observed interference effects of quantum states), and 2) the amplitude for the state |ψ> to be in the state |φ> by first going into the state |i> is just $\langle \psi |i\rangle \langle i|\phi \rangle$, which can be proved easily. $\sum \langle \psi |i\rangle \langle i|\phi \rangle$ can be rewritten as $\sum \langle \psi |(|i\rangle \langle i|\phi \rangle )=\langle \psi |\sum |i\rangle \langle i|\phi \rangle$ if we assume some sort of distributive property. This allows us to write $|\phi \rangle =\sum |i\rangle \langle i|\phi \rangle$, and so we enter the domain of the vector space. The only step missing from making this a Hilbert space is to show that <a|a> is real and positive...but this implies $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$, which I can't for the life of me see why it would be true. I hope this gives you some idea about how I'm thinking about this problem. In effect, we don't need Hilbert spaces to do quantum mechanics. 76.68.247.201 (talk) 02:41, 26 October 2010 (UTC)

I concur with BenRG that this is a time-reversal thing. The $\langle \phi |\psi \rangle ^{*}=\langle \psi |\phi \rangle$ is part of the definition of the inner product of a Hilbert space, which existed as a concept before it was ever used to model quantum mechanics. One of the reasons that representing quantum states as vectors in a Hilbert space works well is because the way the inner product of a Hilbert space is defined results in an automatic invariance of joint probability densities under time reversal. There's a time order dependence to the meaning of $\langle \phi |\psi \rangle$. That expression is the amplitude that the system is in the state $\psi$ as considered at some earlier time, and is in the state $\phi$ as considered at some later time. $\langle \psi |\phi \rangle$ is then the time reversal of that evolution of the system state, in that it's the amplitude that the system is in the state $\phi$ as considered at some earlier time, and is in the state $\psi$ as considered at some later time. For time reversal symmetry to hold, we require that the probability that that combination of before-and-after states occurs is the same, whether looking at the system forward in time or backward in time, i.e., we require that $\left|\langle \phi |\psi \rangle \right|^{2}=\left|\langle \psi |\phi \rangle \right|^{2}$, which happens automatically due to how the inner product is defined. The above explanation may make more sense if you think of the state as being the superposition of various state vectors, at both the earlier time and the later time.

There's actually a complication to the above in that a state function is actually complex conjugated under time reversal. I.e., the time-reversal transformation of $\psi (r,t)$ is $\psi (r,t)\to \psi (r,-t)^{*}$, where * means complex conjugation, not Hermitian conjugation. However, that doesn't affect the basic argument above because $\left|\langle \phi ^{*}|\psi ^{*}\rangle \right|^{2}=\left|\langle \phi |\psi \rangle ^{*}\right|^{2}=\left|\langle \phi |\psi \rangle \right|^{2}$.

I don't know how weak interactions fit into the above.

As a caveat, I'm just basing the above on my own understanding; it isn't something I got out of a quantum mechanics textbook. The QM textbooks I have generally effectively just say "quantum states are modeled with vectors in a Hilbert space", without saying anything about why quantum states are modeled that way. Hopefully bigger physics heavyweights like BenRG will shred my above post if my reasoning is way off. Red Act (talk) 03:27, 26 October 2010 (UTC)

That makes sense, thanks a lot 76.68.247.201 (talk) 04:32, 26 October 2010 (UTC)

Making Helium

If we're running out of helium, why not plunk down a bunch of radioactive waste in (well-sealed) room full of air (or some gas). Unless I'm missing something, the beta and alpha particles it gives off should slow down and then combine into helium... that's how the helium we get out of natural gas pockets forms, right? The only problem I could see would be that the room would potentially need to be REALLY big, and a lot of the walls would get irradiated... but surely if helium becomes valuable enough these aren't insurmountable costs. 96.246.58.133 (talk) 20:53, 25 October 2010 (UTC)

No need for a big room, nuclear reactors do this every day. I'm not sure if they collect the helium or simply vent it, but they definitely do need to handle it. But, I don't think they make enough to replace our uses of it. Maybe someone has time to do some order of magnitude calculations? Ariel. (talk) 21:07, 25 October 2010 (UTC)

The total production in the Earth by all the radioactivity all the way in to the centre is less than 10 000 ton/year our consumption was 12 000 ton/year in 1996 CE. We can not collect all the radioactivity in the Earth in a room and not prevent most of the helium from escaping the Earth.

Derivation:

Total amount of geothermal power, 30 TW (Twice the total human energy consumption so nuclear power will produce no significant amount of He)

A typical alpha decay energy, 4 MeV (Uranium, Thorium)

Molar weight of He 4 g/mol

1/mol =1/6e23

If we put it together we get: 30e12 W/(4 MeV)*(1 mol/6e23)*4g/mol*1 year = 9850 ton (Google calculator[4] understands the formula.) —Preceding unsigned comment added by Gr8xoz (talk • contribs) 22:14, 25 October 2010 (UTC)

There's more discussion of the problem here in the Reference Desk archives (from last month). Physchim62 (talk) 02:24, 26 October 2010 (UTC)

We could always just run out to Jupiter to get more if it truly is that important. Googlemeister (talk) 15:11, 26 October 2010 (UTC)

Golden cross on the Moon

Hello! I'm currently transcribing my great-great-great aunt's memoirs, and she mentions a strange phenomenon that took place on a Good Friday, circa 1890. She describes it thus: "at Easter the moon is always at the full, and right across the moon was a beautiful cross in vivid gold [...] We hurried home to be in time to tell our parents & they came out to look, but by that time one part of it had faded. Thus † became [right half of a cross]." Does anyone have a scientific explanation for this event?--Midgrid(talk) 20:54, 25 October 2010 (UTC)

I should probably point out that this event was also reported in the local newspapers at the time, and was witnessed by other people.--Midgrid(talk) 21:47, 25 October 2010 (UTC)

One obvious possibility would be a cross-shaped cloud at sunset that happened to be in front of the moon. If a group of people who happen to be at the same place at the same time point it out to each other after one of them notices, then there would be multiple witnesses. And I totally could see one of those little news-starved small town newsletters printing something as minimally earth-shattering as that. Decades ago, I visited my girlfriend at the time and her parents overnight in the tiny town she grew up in, and that event was big enough news that it wound up in the local newspaper. Red Act (talk) 22:01, 25 October 2010 (UTC)

I assume people could recognize any ordinary cloud, but there are a long list of oddities under the general heading of "optical phenomenon" which are quite rare in many areas and very striking to observe, which are produced by a relatively transparent layer of ice crystals. For example a moon dog produces extra spots that can be part of a cross-like pattern, though the ring would seem jarring to someone expecting a +-shaped cross - still, some religious traditions use a cross with a broad ring, so that might not disqualify it. The article sun dog gives a broader variety of how such things could look. Wnt (talk) 22:11, 25 October 2010 (UTC)

See this page for a description of a similar event. Looie496 (talk) 23:06, 25 October 2010 (UTC)

Comet identification

From the same source as my above question, my relative mentions a bright comet visible in the sky, circa 1879. Can anyone identify a possible candidate for me?--Midgrid(talk) 20:57, 25 October 2010 (UTC)

Googling comet 1879 yields much junk but mentions several comets, including 5D/Brorsen, which, the article states, was visible for 4 months. Comet Tuttle (talk) 21:07, 25 October 2010 (UTC)

(ec) Many comets were visible from Earth in 1879, and if your "circa" is interpreted as "plus or minus a few years," the list grows even larger. Tempel 1 is one possibility; several small comets are known only as "Comet 1879a, 1879b, 1879c", and so on. A letter to Nature dated July 3, 1879, describes astronomical details for one comet observed in early summer. Nimur (talk) 21:08, 25 October 2010 (UTC)

Thanks, but it's almost certainly not that one, as from the letter I gather that one needed a telescope to see it, whereas the one my relative mentions was clearly visible to the naked eye.--Midgrid(talk) 21:31, 25 October 2010 (UTC)

We certainly got our vengeance on Tempel 1, though it took a while. Comet Tuttle (talk) 21:11, 25 October 2010 (UTC)

Brorsen sounds like a likely possibility, as it passed close the to the Earth and my relative writes that the comet she saw was very bright and had an extremely long tail. "Circa" is not a big variation in this case: she was born in 1870 (or possibly 1871) and says she was "about nine" when she saw the comet. Curiously, she goes on to say that she doesn't think a comet has been visible since (writing c. 1935), despite the fact that even Halley's Comet, the most famous of its kind, passed by in the intervening period!--Midgrid(talk) 21:30, 25 October 2010 (UTC)

Not only that, but the Great January comet of 1910 was the brightest comet of the century. No offense, but you're talking about a person who said she saw a golden cross on the Moon on Easter. Her reliability with regard to astronomical observations, I would say, is low. Comet Tuttle (talk) 21:42, 25 October 2010 (UTC)

No offence taken. :) I did initially think that she might have confused the 1879 comet with Halley, but there was such a large time difference between them.--Midgrid(talk) 21:49, 25 October 2010 (UTC)

Pulpit Rock

would it hurt to fall from Pulpit Rock

http://www.youtube.com/watch?v=J6OtiTrPk80 —Preceding unsigned comment added by Kj650 (talk • contribs) 21:17, 25 October 2010 (UTC)

Don't troll the reference desk. Theresa Knott | Hasten to trek 21:23, 25 October 2010 (UTC)

To fall does not hurt but hitting a rock or the ground may be hurt but I think there are some probability that the brain is destroyed before you feel any pain, are here anyone that knows more about neurology than me that can answer that? --Gr8xoz (talk) 22:21, 25 October 2010 (UTC)

The only conclusive way to answer such a thing is to do an experiment, e.g. a fMRI scanner is rigged to a subject tossed out of a helicopter... I don't think it's been done, though I haven't gotten through to Wikileaks lately. Note that an action potential in long myelinated axons in vertebrates can get up to 110 m/s or greater, which I think is 250 mph, versus the figure given for skydivers at terminal velocity of 120 mph. Which means that the hapless jumper landing feet first after falling prone should have 2 m = ~ 1/30 s to appreciate the sensation, minus the time needed to send and receive the signal. Note however that subjective time behaves strangely even under the mild alterations imposed by psychoactive substances, and I would not want to make any assumptions about how long the last fraction of a second might last under those conditions. Please note that Wikipedia does not give medical advice and disclaims all responsibility for any conclusions you might draw, in this world and the next. Wnt (talk) 22:42, 25 October 2010 (UTC)

On a recent NGC channel documentary they showed how you can jump safely from a height of about 10 meters on a hard concrete surface. You have to make sure you have enough forward velocity and make yourself small and roll when you hit the ground. Count Iblis (talk) 01:09, 26 October 2010 (UTC)

Here's an 8 sec clip of someone doing a parkour jump of 25 feet: [5]. (You would not have to worry about landing technique after the 600 m fall from Pulpit Rock). WikiDao ☯ (talk) 02:04, 26 October 2010 (UTC)

The classic response is - no, it won't hurt at all to fall from Pulpit Rock, but it will sting a bit when you hit the ground. Richard Avery (talk) 08:01, 26 October 2010 (UTC)

But one might aim for the water not the ground. This is a discussion about the question What is the maximum height from which a person can make a free dive without incurring physical injury?. A useful factoid observed is that Feet-first impact can be lethal due to explosive rupture of the large colon. Presumably that is not a pleasant experience. Cuddlyable3 (talk) 10:51, 26 October 2010 (UTC)

It wouldn't sting if you fell with a parachute on your back, base jump style, or with a wing suit on. As my physics teacher used to say referring to car crashes (and this is about all I remember) "It's not the crash that hurts, it's the sudden deceleration". Smartse (talk) 11:00, 26 October 2010 (UTC)

Having finally watched that YouTube video, I have to say that the idyllicness was somewaht spoiled by all the human chatter in the background. Maybe that's what made our poster think of jumping. HiLo48 (talk) 08:34, 26 October 2010 (UTC)

Pushing would also work. Cuddlyable3 (talk) 10:53, 26 October 2010 (UTC)

I doubt that. Humans have the tendency to get quite excited and make a lot of noise when you start pushing people off very high cliffs Nil Einne (talk) 11:24, 26 October 2010 (UTC)

Their noise recedes. See Doppler effect. Cuddlyable3 (talk) 20:19, 26 October 2010 (UTC)

Only of those you successfully push. Under the conditions where there are plenty of people or when other people are likely to see what you do as seems likely in the above case, I stand by my comment. Nil Einne (talk) 23:25, 26 October 2010 (UTC)

I think the pretty blond could easily convince me not to jump... but maybe that's just me... Dismas|^(talk) 03:44, 27 October 2010 (UTC)

Are you sure she'll try? She may be afraid you're Cuddlyable3... ;-) P.S. To avoid confusion make sure you read the responses above this. Nil Einne (talk) 14:56, 28 October 2010 (UTC)

There's no confusion. Dismas is the one prepared for a pole jump. Cuddlyable3 (talk) 20:33, 28 October 2010 (UTC)