Wikipedia:Reference desk/Archives/Science/2008 March 12

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March 12

vitamin A

wat way can i chemicaly make or extract pure Vitamin A. And if i can't get it pure, how can i get it the purest? —Preceding unsigned comment added by 76.14.124.175 (talk) 04:03, 12 March 2008 (UTC)

Probably easy to extract it from a commercial over-the-counter vitamin supplement. Depends what you want to do with it how pure is "pure". DMacks (talk) 06:35, 12 March 2008 (UTC)

Vitamin A can be make synthetically - you might need to know a lot of chemistry though - a good library should have something about it's synthesis in the organic chemistyr section.

It's possible to buy vit A (Retinol) from a chemical supplier at 99% pure.

It would also be possible to extract vitamin A from a sample containing it - though you would probably need to do chromatography to get it pure.87.102.17.32 (talk) 13:31, 12 March 2008 (UTC)

Just a reminder that does of vitamins much above the recommended make you ill or worse. I have stopped taking multivitamins after reading scientic research which showed that they do more harm than good, increasing cancer rates and reducing survival and recovery. (I do take Vitamin D though). Another example: explorers have died from eating liver from polar bears or huskies, which contain poisonous quantities of Vitamin A. 80.0.108.245 (talk) 19:51, 16 March 2008 (UTC)

CHloroacetic acid

Below is how it is made. But wat are the exacts? do i have to heat it? do i have to use pure chlorine? how do i use the catalyst?

Chloroacetic acid is synthesized by chlorinating of acetic acid in the presence of red phosphorus, sulfur, or iodine as a catalyst:

CH3CO2H + Cl2 → ClCH2CO2H + HCl —Preceding unsigned comment added by 76.14.124.175 (talk) 04:32, 12 March 2008 (UTC)

I'm gonna go with "if you don't know how to look these kinds of things up in the chemistry literature, you shouldn't be doing this reaction". DMacks (talk) 06:33, 12 March 2008 (UTC)

Chemisrty literature??????????????? —Preceding unsigned comment added by 76.14.124.175 (talk) 23:48, 12 March 2008 (UTC)

chemotherapy or radiation therapy for obesity

Could anticancer therapies be tweaked or redesigned to attack fat cells? Thanks, 05:37, 12 March 2008 (UTC)Rich (talk) 05:38, 12 March 2008 (UTC)

Have you read our article on antineoplastics and/or chemotherapy? There are a number of different techniques, but they typically rely on cancer's fast growth rate (for example, they slow/damage new DNA synthesis). This works really well on fast growing cells (such as malignant tumors and hair follicles), it doesn't typically affect slow growing cells (such as neurons and adipocytes (aka fat cells)). So, generally speaking, I'd say no. (EhJJ)^TALK 11:43, 12 March 2008 (UTC)

Also, it would be bad to get rid of fat cells. You want a lot of small fat cells for smooth skin. With no fat cells you would look like a skeleton, and with a few fat cells they would tend to become large and clump together, giving you a lumpy appearance. StuRat (talk) 15:01, 12 March 2008 (UTC)

Good answers, thanks.Rich (talk) 21:09, 12 March 2008 (UTC)

steroids

Whats the difference between anabolics and steroids? —Preceding unsigned comment added by 75.4.67.159 (talk) 05:43, 12 March 2008 (UTC)

Our anabolic steroid page is a good place to start. DMacks (talk) 06:11, 12 March 2008 (UTC)

Quantum Mechanics: Entangled Wave Function

The equation 9, or (EPR9) for short here, in the original paper of EPR paradox gives a wave function of two entangled particles

$\Psi (x_{1},x_{2})=\int _{-\infty }^{\infty }e^{(2\pi i/h)(x_{1}-x_{2}+x_{0})p}dp$

(EPR9)

where $h$ is Planck's constant, $x_{1}$ and $x_{2}$ are the variables describing the two particles and $x_{0}$ is just some constant. According to reduction of the wave packet, when an observable $B$ of the first particle is measured, (EPR9) can be expanded by the eigenfunctions $v_{1}(x_{1}),v_{2}(x_{1}),v_{3}(x_{1}),...$ of $B$ in the form

$\Psi (x_{1},x_{2})=\sum _{s=1}^{\infty }\varphi _{s}(x_{2})v_{s}(x_{1})$

(EPR8)

where $\varphi _{1}(x_{2}),\varphi _{2}(x_{2}),\varphi _{3}(x_{2}),...$ are the corresponding coefficients to the eignefunctions. If $B$ is a continuous observable, the coordinate of the first particle, (EPR8) can be written as

$\Psi (x_{1},x_{2})=\int _{-\infty }^{\infty }\varphi _{x}(x_{2})v_{x}(x_{1})dx$

(EPR15)

According to the paper, the eigenfunctions of $B$ is

$v_{x}(x_{1})=\delta (x_{1}-x)$

(EPR14)

which has corresponding eigenvalue $x$. The first question is how come the eignefunction and the eigenvalue of $B$ are (EPR14) and $x$, respectively? It seems that

$B=x_{1}$

(1)

and if we let

$f(x_{1})=x_{1}-x$

(2)

then

$v_{x}(x_{1})=\delta (x_{1}-x)=\delta (f(x_{1}))$

(3)

Find the solution of

$f(x_{1})=0$

(4)

we have

$x_{1}-x=0$

(5)

$x_{1}=x$

(6)

the right-hand side of (6) is the eigenvalue of $B$. Similarly, the eigenvalue of the observable

$Q=x_{2}$

(EPR17)

can be found by knowing

$\varphi _{x}(x_{2})=\int _{-\infty }^{\infty }e^{(2\pi i/h)(x-x_{2}+x_{0})p}dp=h\delta (x-x_{2}+x_{0})$

(EPR16)

and let

$g(x_{2})=x-x_{2}+x_{0}$

(7)

The solution of

$g(x_{2})=0$

(8)

is

$x_{2}=x+x_{0}$

(9)

Again, the right-hand side of (9) is the eigenvalue of $Q$ which complies with the paper. But it still doesn't explain how to figure out the eignefunction (EPR14).

To continue the unsolved discussion last time, the second question is how to denote the entangled wave function (EPR9) in bra-ket notation? If it can be done, it should help with respect to the last discussion. The bra-ket notation of (EPR9) is supposed to be in the Hilbert space which is the tensor product of the state spaces associated with the the two particles. - Justin545 (talk) 06:36, 12 March 2008 (UTC)

Hi, I'm sorry I haven't followed up to the old thread yet, but maybe a response here will serve the same purpose.

There are many ways to write (EPR9) in bra-ket notation; for example I could just write $|\Psi \rangle$ where Ψ is defined by (EPR9). In terms of tensor products of kets inhabiting the state spaces of the individual particles, I could write for example $\Psi =\int _{-\infty }^{\infty }|x\rangle _{1}\,|x+x_{0}\rangle _{2}\,dx=\int _{-\infty }^{\infty }e^{(2\pi i/h)x_{0}p}\,|p\rangle _{1}\,|{-}p\rangle _{2}\,dp$. I'm not sure those are properly normalized, to the extent that these mathematical monstrosities can be considered to be normalized to begin with. The product $|a\rangle |b\rangle$ might also equivalently be written $|a\rangle \otimes |b\rangle$ or $|a,b\rangle$ or $|ab\rangle$. The subscripts 1 and 2 just indicate which subspaces the kets inhabit; they could be left off since the two subspaces are isomorphic in this case.

I'm not sure I understand your first question. Finding eigenfunctions of the position operator in a single-particle space involves solving equations of the form $B\Psi =x\Psi$ where B(z) = z and BΨ is a pointwise function product. It should be clear enough that the only possibilities for Ψ here are functions that are zero everywhere except at a point, and the "normalized" versions of these functions are the delta functions, which form an orthonormal eigenbasis. In the two-particle space things are a bit more interesting. You're now solving $B\Psi =x\Psi$ where $B(z_{1},z_{2})=z_{1}$. The normalized solutions here are $\Psi _{x}(x_{1},x_{2})=\delta (x-x_{1})g(x_{2})$ where g is any normalized function of x₂. These do not form a basis; there are far too many of them for that. You have to choose arbitrarily some orthonormal basis for the functions g. This happens because there are degenerate eigenvalues; the discrete analogy is that there's only one orthonormal eigenbasis for diag(1,2,3) but many for diag(1,1,2). -- BenRG (talk) 12:55, 12 March 2008 (UTC)

It's reasonable making the ket be the function of the corresponding eigenvalue since each eigenvalue identifies an unique basis or eigenfunction. But, I am a bit confused with the bra-ket notation $\Psi =\int _{-\infty }^{\infty }|x\rangle _{1}|x+x_{0}\rangle _{2}\,dx$ since I expect the bra-ket notation should be in the form

$\Psi =c_{1}|a_{1}\rangle _{1}|b_{1}\rangle _{2}+c_{2}|a_{2}\rangle _{1}|b_{2}\rangle _{2}+c_{3}|a_{3}\rangle _{1}|b_{3}\rangle _{2}+...$

(10)

rather than in the form

$\Psi =\int _{-\infty }^{\infty }\left(c_{1}|a_{1}\rangle _{1}|b_{1}\rangle _{2}+c_{2}|a_{2}\rangle _{1}|b_{2}\rangle _{2}+c_{3}|a_{3}\rangle _{1}|b_{3}\rangle _{2}+...\right)dx$

(11)

It seems the integral $\int _{-\infty }^{\infty }\cdot \,dx$ surrounding the ket can not be removed. But, will the integral of the ket yield another "ket" in the same space? Another confusion is about the momentum part of the bra-ket example $\Psi =\int _{-\infty }^{\infty }e^{(2\pi i/h)x_{0}p}|p\rangle _{1}|{-}p\rangle _{2}\,dp$. I am not able to figure out $e^{(2\pi i/h)x_{0}p}$ in it.

Apologies for obscuring my first question. My first question is just to understand why the eigenfunction of $B$ is a "delta function". Just wonder how the delta function (EPR14) is mathmatically derived. As you said "Finding eigenfunctions of the position operator in a single-particle space involves solving equations of the form $B\Psi =x\Psi$ where B(z) = z and BΨ is a pointwise function product." But I can not understand why it's pointwise. Excuse my poor quantum mechanics, I left so many question marks here :-) Justin545 (talk) 08:43, 13 March 2008 (UTC)

The integral is the sum, it just happens to be a sum with uncountably many terms. You need an uncountable sum here because the wave function is a superposition of uncountably many tensor-product states—the particles could be at x and x+x₀ for any real x. I picked somewhat arbitrarily the position basis vectors $|x\rangle (x')=\delta (x-x')$ and the momentum basis vectors $|p\rangle (x)=e^{(2\pi i/h)xp}$. They're somewhat arbitrary because they're only unique up to scalar multiplication, but they're eigenvectors of the appropriate operators with the appropriate eigenvalues (unless I got the sign convention backwards). Then $|x_{1}\rangle |x_{2}\rangle (x_{1}',x_{2}')=\delta (x_{1}-x_{1}')\delta (x_{2}-x_{2}')=\delta ((x_{1},x_{2})-(x_{1}',x_{2}'))$ and $|p_{1}\rangle |p_{2}\rangle (x_{1},x_{2})=e^{(2\pi i/h)(x_{1}p_{1}+x_{2}p_{2})}$. So in particular $e^{(2\pi i/h)x_{0}p}|p\rangle \,|{-}p\rangle (x_{1},x_{2})=e^{(2\pi i/h)(x_{1}-x_{2}+x_{0})p}$, which is where my momentum integral form came from. The position integral one is odder. When $x_{1}-x_{2}+x_{0}\neq 0$, EPR9 gives $\int _{-\infty }^{\infty }e^{i({\text{some nonzero real}})p}\,dp$, which to a mathematician is undefined but to a physicist is zero. When $x_{1}-x_{2}+x_{0}=0$, EPR9 gives $\int _{-\infty }^{\infty }dp$, which to a physicist is the peak of a delta function. So EPR9 describes a "function" that's zero everywhere except on the line $x_{1}-x_{2}+x_{0}=0$ where it's infinity, and my position integral expressed that more directly.

Let me explain the operators in a finite-dimensional case. Let's say we have a four-state system with position states $\left\{\left({\begin{array}{r}1\\0\\0\\0\end{array}}\right),\left({\begin{array}{r}0\\1\\0\\0\end{array}}\right),\left({\begin{array}{r}0\\0\\1\\0\end{array}}\right),\left({\begin{array}{r}0\\0\\0\\1\end{array}}\right)\right\}$ and momentum states ${\frac {1}{2}}\left\{\left({\begin{array}{r}1\\1\\1\\1\end{array}}\right),\left({\begin{array}{r}1\\i\\-1\\-i\end{array}}\right),\left({\begin{array}{r}1\\-1\\1\\-1\end{array}}\right),\left({\begin{array}{r}1\\-i\\-1\\i\end{array}}\right)\right\}$ (the Fourier basis). We can arbitrarily assign a distinct real number to each position and to each momentum. Say the positions are 1,2,3,4 and the momenta are 0,1,2,−1. Then there exists a matrix which scales each position/momentum axis by the corresponding real number. For the position basis it's just diag(1,2,3,4), while for the momentum basis it's U diag(0,1,2,−1) U⁻¹, where U is the unitary matrix whose columns are the aforementioned Fourier basis vectors. This matrix will always be Hermitian (it's a theorem that a matrix is Hermitian if and only if it can be written in the form UAU⁻¹ where U is unitary and A is real diagonal). In this case the scaling factors were all distinct, so by solving the eigenvalue equation we can recover the original basis from the Hermitian matrix. If some scaling factors are equal then all you can tell is that a particular (hyper)plane was scaled by that factor; you can't uniquely recover the basis vectors lying in that hyperplane. That's the case for a four-state system that's the product of two single-particle two-state systems, where the position of the two particles might be represented by the matrices diag(1,2,1,2) and diag(3,3,4,4) respectively. In the continuous case you can't write down matrices any more, but the differential operators serve the same purpose. In order to get the right eigenbasis and eigenvalues, the position operator has to multiply the wave function by a real number corresponding to the position, which is why I described it as a pointwise function product. It might have been better to say that B is an operator defined by (B f)(x) = x f(x). -- BenRG (talk) 14:32, 14 March 2008 (UTC)

Cancer

Why do swelling often occurs around the tumor? there doesn't seem to be an answer in wikipedia.

• this is not homework —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 06:52, 12 March 2008 (UTC)

There's no single answer; tumor can block lymphatic drainage and cause fluid accumulation; cytokines secreted by tumor cells can cause inflammation; some tumors upregulate angiotensin II type 1 receptors, triggering chronic inflammatory response; some tumors cause the formation of new blood vessels (angioneogenesis) causing hyperemia and swelling. - Nunh-huh 06:59, 12 March 2008 (UTC)

And then there's the simple method of the tumor growth itself pushing on the surronding tissue, which then expands outward, similar to how a growing weed can push asphalt upward. StuRat (talk) 14:57, 12 March 2008 (UTC)

Tumors often elicit immune responses from cytotoxic T lymphocytes, but also bear in mind that tumors promote angiogenesis as well. Wisdom89 _{(T / ^C)} 18:49, 12 March 2008 (UTC)

Hepatitis C

How dangerous and contagious is hepatitis C, can it be transmitted through a cats claws? —Preceding unsigned comment added by DSTiamat (talk • contribs) 08:31, 12 March 2008 (UTC)

If you'll read our article on Hepatitis C, you'll find that it is spread by blood-to-blood contact, and over 90% of cases are spread through blood products or recreational drug use. A less common form of spread is via sexual contact. Though one can make up scenarios in which an infected person is scratched by a cat, who then immediately scratches and infects someone else, this is not something that's ever been reported as actually happening. Hepatitis C infections can range from asymptomatic to fatal, so obviously its dangerousness varies greatly, depending on a number of other factors, including luck. - Nunh-huh 09:10, 12 March 2008 (UTC)

If you want a disease associated with cat scratches, look up cat scratch fever. StuRat (talk) 14:52, 12 March 2008 (UTC)

Who makes the 120mm fans used on the space shuttle?

I have no Idea on where to go to start looking for such information, how many different manufacturers are used and what they are used for...just think it'd be interesting. —Preceding unsigned comment added by 71.237.205.31 (talk) 09:11, 12 March 2008 (UTC)

You need to start with a specific module so you can narrow down to the country that built the module. Then, you can hope some of the contracts were made public within that country. -- kainaw™ 15:16, 12 March 2008 (UTC)

The shuttle has at least four different types of fans in the environmental control system. There are two cabin fans, six avionics bay cooling fans, three IMU fans, and at least one airlock booster fan; there may be more I'm forgetting. I'm not sure if they are all identical (I would assume not). I don't know any details (or even where to start looking) about make and model of the fans. anonymous6494 18:32, 12 March 2008 (UTC)

Whether you are looking for low noise, high efficiency, or

the security of cooling fans designed and built for the long haul, Howden Cooling Fans has lead the market since 1955. With over 150 years of experience, Howden is the world’s largest and longest established manufacturer of air and gas handling equipment. Today, no one equals Howden’s accumulated knowledge of fans, with applications ranging from cooling fans on the space shuttle to 15,000 hp axial fans for power station boilers.

Biochemistry

The process in purin purification by Molecular exclusion62.24.99.237 (talk) 10:51, 12 March 2008 (UTC)

You probably meen purine and "molecular exclusion" can be searched for on the web, or may be described at Size exclusion chromatography87.102.17.32 (talk) 18:40, 12 March 2008 (UTC)

Could you be more specific87.102.17.32 (talk) 18:41, 12 March 2008 (UTC)

Panic Attack while teaching my kid

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. Please consult a physician, psychiatrist or psychologist--(EhJJ)^TALK 11:33, 12 March 2008 (UTC) (EhJJ)^TALK 11:33, 12 March 2008 (UTC)

heat conductor

Why does the a plain cloth catches fire when put near a flame at about 10 cm compared to the cloth wraping a coin in it? —Preceding unsigned comment added by 165.21.155.69 (talk) 12:43, 12 March 2008 (UTC)

Maybe it gets hotter quicker without the coin?87.102.17.32 (talk) 13:33, 12 March 2008 (UTC)

Yes, the coin is a good thermal conductor, so pulls heat out of the cloth, acting as a heat sink, until it gets up to temp, then it will no longer draw off heat. StuRat (talk) 14:49, 12 March 2008 (UTC)

we are looking for info regarding mining

What is fools gold ( piriet) and why do they call it fools gold?

The impact that the mining of gold has on SA's environment: The use of sinade and uranium as by-products in the mining process The influence of mining on water and the effect of water on buildings Mine heaps amounts of rock removed from the earth surface. Air Pollution. —Preceding unsigned comment added by 196.211.98.154 (talk) 13:03, 12 March 2008 (UTC)

Fools gold is pyrites (spelling) it's actually Iron Sulphide.

sinade = cyanide which is used to extract the gold.87.102.17.32 (talk) 13:13, 12 March 2008 (UTC)

Gold mining should be of interest. there's some info on the enviromental hazards of cynaide seeCyanide_process#Effects_on_the_environment.87.102.17.32 (talk) 13:16, 12 March 2008 (UTC)

Also try http://www.google.co.uk/search?hl=en&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=gold+mining+environment&spell=1

Looks like uranium is a by product and a pollutant produced when gold in mined http://www.ingentaconnect.com/content/klu/gejo/2004/00000061/00000002/00002867;jsessionid=1rj6eslojhj0m.alexandra? —Preceding unsigned comment added by 87.102.17.32 (talk) 13:20, 12 March 2008 (UTC)

Looks like South African homework. :-P --98.217.18.109 (talk) 14:29, 12 March 2008 (UTC)

Hi. It might be. OP is from South Africa. ~AH1^(TCU) 00:48, 13 March 2008 (UTC)

Fool's Gold has the chemical formula FeS2. The name is derive from it's physical appearance. Miners mining for gold occasionally stumble upon shiny flaky materials mistaking it for actual gold. Hence the term fool's gold. Hey mrs tee (talk) 09:30, 19 March 2008 (UTC)

Cuttlefish wants to cuddle

How do Cuttlefish know how to blend in with a given environment if they can't see in color? Why would their skin have such a wide range of colors possible if they can't distinguish between most of them? --98.217.18.109 (talk) 15:07, 12 March 2008 (UTC)

According to our article, cuttlefish change color to camouflage themselves as well as communicate. The camouflage function would benefit from a range of colors even if they are colors the cuttlefish can't themselves see. As for how they match colors they can't see, I don't know how they do that, and according to this article, at least one cuttlefish camouflage expert doesn't know either. --Allen (talk) 15:33, 12 March 2008 (UTC)

This article is also fascinating. —Keenan Pepper 18:37, 12 March 2008 (UTC)

See also chromatophore, specifically the section on Cephalopod chromatophores. Rockpocket 19:00, 12 March 2008 (UTC)

Let's get this straight -cuttlefish are colour blind but they do match themselves to the colour of the surroundings??? so they camoflage based on the grey scale image they see —Preceding unsigned comment added by 87.102.17.32 (talk) 20:55, 12 March 2008 (UTC)

Well, you're assuming that the color change is in response to a visual signal. Maybe it's not; maybe something in the cuttlefish's skin (if that's the right word) responds directly to the color of the light hitting it, without going through the eye at all. --Trovatore (talk) 21:04, 12 March 2008 (UTC)

I wondered about light sensors in the skin..

Though in one of the references above -(nytimes) - it describes the placement of a high contrast pebble in the fish tank, the cuttlefish then 'looking at it' and then adding a black spot to its skin pattern.. That shows (I think) at least that the eyes are at least a part of the camoflage set-up..87.102.17.32 (talk) 21:50, 12 March 2008 (UTC)

I hadn't read the Science article when I gave my answer; now I've scanned it and it looks like the NYT reporter Carl Zimmer must have made an error. He quotes Dr. Hanlon as saying "we don't know how" [in Zimmer's words:] "they see a world without color, but their skin changes rapidly to any hue in the rainbow." But the Science article clearly demonstrates that cuttlefish's skin does not change to any hue in the rainbow. So maybe there's no mystery: cuttlefish are color blind, and therefore have trouble camouflaging themselves against certain colorful backgrounds. But on the other hand, there is the last sentence of the Science paper: "However, the vexing question of how S. officinalis masters the task of camouflage in chromatically rich environments, such as those found at shallow depths of water, remains to be answered." So Mathger et al. are saying there is a mystery... but what are they referring to? The paper doesn't seem to cite any instances of S. officinalis mastering the task of camouflage in any truly chromatically rich environment. At least not that I can see. --Allen (talk) 02:21, 13 March 2008 (UTC)

urea formation

Hi all read the article on urea cycle found it a bit confusing... could anyone tell me how the NH2 group removed from an amino acid in deamination is converted to ammonia (NH3), Where does this hydrogen come from? I believe the NH2 is converted to NH3 before it reacts with CO2 to form urea, but please correct me if that is wrong. Thanks. —Preceding unsigned comment added by 172.142.47.24 (talk) 21:10, 12 March 2008 (UTC)

Are you refering to "conversion of glutamate to ammonium and α-ketoglutarate." - the 'co-factor' is NAD+ see Glutamate dehydrogenase. Was that the bit you meant. the amino acid is oxidised to a keto compound.

In the big diagram http://en.wikipedia.org/wiki/Image:Urea_cycle_2.png 'ammonia' remains bound to the molecule as an iminium cation, with loss of fumarate, water hydrolysis the iminium cation to urea..87.102.17.32 (talk) 21:42, 12 March 2008 (UTC)

This is probably the easier diagram - if you are still stuck can you reference your questions to this diagram. Just to make it easier (assuming it's relevent)87.102.17.32 (talk) 21:45, 12 March 2008 (UTC)

The "fish scales" in the clouds are back!

pink-teal iridescence, traces of hints of blue, violet, orange

pink-teal iridescence, orange-red halo, blue, green, and violet patches, blue-green inner halo glow, near-sun whiteness

Hi. Remember my previous question? The one a few weeks ago with the 8 deg halo and the iridescent "fish scales" in the clouds? Well, here's the story. Yesterday I saw a roughly 25 deg from the sun rainbowlike halo, and a sundog while the sun was in a cirrus-cirrostratus-altostratus cloud. Anyway, remember last time I saw those "fish scales" in the clouds and that halo as I was typing? Well, I didn't have a camera. Well, guess what? This time I was in a car. I saw it again. At first I thought we didn't bring the camera, but it turns out we did. Yay! I took dozens of pictures, but erased every one of them except seven, because the others were too indistinct. Anyway, I uploaded two of them. Click on the images if you wish. Now, remember last time there was a cirrocumulus cloud passing by as I saw the iridescent fish scales? This time, cloud after cloud passed by or near the sun. The way I saw them were much, much more colourful than in the pictures, probably because the camera was automaticly set to neutral colour or something. Anyway, the clouds were probably stratocumulus, but some of them were very fibrous, and there were some clouds slightly higher than others. Look carefully for the colours, because they're much harder to see in the images than in real life, which is why I only picked the best two. Notice in the pictures: The organge halo, the pink-teal iridescence, and a hint of rainbow colours. What do you think they are, and do we have an article on them? Thanks. ~AH1^(TCU) 23:23, 12 March 2008 (UTC)

It's not exactly a glory, but that page may help. Cheers Geologyguy (talk) 01:01, 13 March 2008 (UTC)

No, it's not a glory; I guess the closest article would be Cloud iridescence. It's simply dispersion of sunlight by water droplets in the clouds. FiggyBee (talk) 03:43, 13 March 2008 (UTC)

There's a nice sample here[1] at the Cloud Appreciation Society site, but I like your "fish scales" description. Your pics wouold go well in the iridescence article to relieve the high contrast ones it has, to show some alternatives. I've seen what you describe – glad to know it's a phenomenon. Julia Rossi (talk) 07:24, 13 March 2008 (UTC)

Wow a RARE PHENOMENON!!! I saw this phenomenon twice and I had my camera one time and uploaded it!!! YAY!!! ~AH1^(TCU) 17:17, 13 March 2008 (UTC)

You might check out Marcel Minnaert's book: The Nature of Light and Color in the Open Air. --Carnildo (talk) 21:40, 13 March 2008 (UTC)