Science desk
< May 31	<< May \| June \| Jul >>	June 2 >

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

June 1

Apparent anomaly

In this article, it states that an excess of bromine prevents the formation of phosphorus pentabromide. How can that be? --Chemicalinterest (talk) 00:46, 1 June 2011 (UTC)[reply]

There's a reference in that article for that statement. Did you follow it? --Jayron32 02:58, 1 June 2011 (UTC)[reply]

The reference is not publicly available. By the way, it says the opposite here. --Chemicalinterest (talk) 12:42, 2 June 2011 (UTC)[reply]

That's a clone of an older version of our own page:) I have access to the cited article, which states "a slight excess of...phosphorus", so I undid this unexplained change. DMacks (talk) 15:46, 2 June 2011 (UTC)[reply]

I can't find a good ref for modern preparation of Phosphorus pentabromide, but did find a ref that the obvious one suggested here tends to lead to PBr7 (!) at least under some conditions. DMacks (talk) 16:00, 2 June 2011 (UTC)[reply]

So it was one of those random fact changers changing the article. PBr5 is more likely to form in high bromine concentration than PBr3 for obvious reasons. --98.221.179.18 (talk) 19:40, 2 June 2011 (UTC)[reply]

What are the black cylinders hanging from the wires across telephone poles?

No I'm not talking about the 120V/240V transformers. I know what those are and they are usually grey and for power lines. I am wondering what are those cylinders on the lower wires. Usually Telephone and Cable wires. Here's a picture that shows about 4-6 of them.

http://ops.fhwa.dot.gov/publications/telecomm_handbook/images/fig8-5.jpg — Preceding unsigned comment added by 76.184.70.229 (talk) 05:09, 1 June 2011 (UTC)[reply]

I'm not sure what the real name for them is, but "distribution box" might be close. They're places that allow fat cables to be tapped to drop small cable pairs (these days two of them) to individual users. PhGustaf (talk) 05:18, 1 June 2011 (UTC)[reply]

this website has heaps of info, I think the answer is they are amplifiers and splitters. Vespine (talk) 05:27, 1 June 2011 (UTC)[reply]

Breakout boxes. Cuddlyable3 (talk) 12:48, 1 June 2011 (UTC)[reply]

What year is this? (in another year numbering system)

If one does not like "artificial" calendars, one can give the date simply by counting days since the most recent equinox. Today's date (June 1) in such a system would simply be "73" (or "252" if you prefer the September equinox), give or take a day depending on timezone.

I do not understand why pretty much all year-numbering systems are so anthropocentric and "artificial". The Jalaali calendar at least tries not to be so artificial in its days and months, but the year numbering is rather religious and anthropocentric.

Year numbering systems need not be so anthropocentric; just as I counted days of the year from an equinox, one can count years of the Great Year from a year whose winter "half" and summer "half" are of equal length. So... using this system, today (2011-06-01) is day 73 of year ???? — Preceding unsigned comment added by 75.40.137.207 (talk) 06:17, 1 June 2011 (UTC)[reply]

According top our article, "In reality the exact duration cannot be given, as the rate of precession is changing over time." So, we need something more precise. Also, the precession of the equinoxes doesn't really have a "start" point from which we can measure. Might I suggest an astronomic phenomenon for which we do have a precise date ? How about the supernova SN 1006, first observed on Earth in the year 1006 AD (under our present deficient calendar system) ? StuRat (talk) 07:32, 1 June 2011 (UTC)[reply]

If I understand your question, you want to know when the last time an equinox occurred at perihelion? This would correspond to Spring + Summer having the same duration as Fall + Winter. According to the orbital dynamics reference model I have on my computer, that last occurred in the year 4025 BC. The reference model is intended to be used for calculating insolation changes over hundreds of thousands of years, so I'm not sure how much precision one should regard that date as actually having. So, assuming I understood your question correctly, the answer is about 6000 years ago. Dragons flight (talk) 08:44, 1 June 2011 (UTC)[reply]

That starts a year count consistent with Young Earth creationism. Cuddlyable3 (talk) 12:43, 1 June 2011 (UTC)[reply]

So it looks like my answer is 6114 (or maybe 6113). Thank you! — Preceding unsigned comment added by 75.40.137.207 (talk) 13:49, 1 June 2011 (UTC)[reply]

... or perhaps I should use 17072. — Preceding unsigned comment added by 75.40.137.207 (talk) 14:00, 1 June 2011 (UTC)[reply]

Since humans invented the calendar and are the only ones who care about it, why wouldn't it be "anthropocentric"? ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:35, 2 June 2011 (UTC)[reply]

Well, if we ever do contact extra-terrestrials, it would be nice to have a way we could communicate dates easily. The Big Bang is the obvious universal event from which to measure dates, but I doubt if we will ever know that precisely enough to use for the basis of a calendar. StuRat (talk) 20:50, 2 June 2011 (UTC)[reply]

That raises the question of what a "day" might be to an E.T. Earth creatures have adapted to the day and year and seasons that Mother Nature has given us. But creatures from a planet with a different length day and year might be totally fish out of water here - just as we would be there. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:17, 2 June 2011 (UTC)[reply]

And using SN 1006 for a reference would only work if their planet was the same distance as earth was from the supernova. Googlemeister (talk) 15:26, 3 June 2011 (UTC)[reply]

True; it might be better to use the time the supernova occurred as measured at the supernova, though the precision of that reference will depend on how closely we – and the aliens – know the distance to the supernova remnant. (Our article gives a distance of "about 7200" light years, which probably means there's about a century of uncertainty in the scale.) Either way, though, these methods rely on the aliens actually having recorded the occurrence of the supernova, and on them maintaining that record (and accurately tracking the time elapsed since); neither assumption guaranteed.

Since in practice one would have to then go on to define some sort of time measurement scale anyway (to describe how long after SN 1006 or another 'baseline' event occurred), it makes much more since to simply express the time and date of historical events in terms of how long before the present message that they occurred. Time units can be expressed as multiples of the frequency of a radio transmission (or interval between bits, or tied to some physical constant). Remember, the SI second is defined in terms of a physical standard which ought to be the same everywhere in the universe.

I'll also note that defining 'simultaneous' events in a galaxy where observers are separated by light years is a non-trivial timekeeping challenge. TenOfAllTrades(talk) 15:58, 3 June 2011 (UTC)[reply]

Kim Robinson's fiction book Red Mars talks about the first group of people who start colonizing Mars. They use some calendar similar to what you propose. If I remember correctly, it's just the angle to the planet from some starting point as seen from the sun, measured in degrees. I don't remember what the starting point is: it could be an equinox (or solistice), or maybe the perihelion or apihelion (the point where the planet is closest or farthest from the sun). – b_jonas 16:18, 5 June 2011 (UTC)[reply]

Pound-mass and pound-force?

Is pound-mass (lbm) a real unit or something derived or obfuscated?

Is it only relevant if you're on the surface of the Earth?

What is the conversion of kg to lbm if you are on the moon?

What is the difference between pound-mass and pound-force (from their respective articles I can only surmise that they are the the same as long as you're on the surface of the Earth)?

What is the difference between pound-mass and slugs? What about "poundal" (before Wikipedia I never heard of this one)?

At uni (aerospace engineering) I was taught to use slugs as imperial mass and treat pounds as a force. Is this right or have I misunderstood?

Anyone answering this please don't treat me like an idiot and please call a spade a spade WP:SPADE. If the answers were obvious there would be no common misconceptions about pounds being a force/mass and how to convert between them etc. Sometimes it just takes someone to explain something in a different way for it to sink in, and for me (and others I think) that explanation hasn't been found.

Relevant articles are Pound-force and Pound-mass. I tried to clear up some of this in the Pound-mass discussion page but was instructed to take my questions here. The Pound-mass article doesn't really mention much about its relationship with the Pound-force or slug (a unit of mass used in aerospace engineering). As an engineer I'm always converting between mass and force, but the conversion table at Pound-force only converts force to force, and my discussion on the Pound-mass talk page led to my shame for converting between units.

Can anyone present real world examples of conversion between these units (mass and force, under influence of gravity of earth and moon just for comparison)?

Thankyou in advance to anyone able to help. — Preceding unsigned comment added by 203.129.23.146 (talk) 09:10, 1 June 2011 (UTC)[reply]

Think of pounds-force and pounds-mass as entirely separate and unrelated units, from a physics point of view, where the ratio between them just happens, by a stunning coincidence, to equal the acceleration of gravity at sea level on the Earth (at some specified latitude I suppose, but I'm not sure which one). That coincidence makes certain calculations easier but has nothing to do with the essence of the units. --Trovatore (talk) 09:20, 1 June 2011 (UTC)[reply]

The weight of an object is different at different locations in the universe. On the other hand, the mass of an object is the same, regardless of its location in the universe. The relationship between lbm, slugs and kg is the same on the Earth, the moon or anywhere else in the universe. It is the relationship between weight on Earth and weight on the moon that is the tricky one. If you treat slugs as imperial mass and pounds as a force you can't go wrong. All engineering students have this difficulty initially, but then it all falls into place when you realise that equations, such as F=ma, can only be correct if you use consistent units. lbm, poundals and ft.s^-2 are consistent units. So too are slugs, lbf and ft.s^-2; and kg, newtons and m.s^-2. Any mixture of those units won't be consistent and so will deliver an incorrect answer. Stick with it. Dolphin (t) 12:16, 1 June 2011 (UTC)[reply]

Imagine a lunar exploration vehicle with a mass of 100 kg. (Its mass is 100 kg on Earth, on the moon, and everywhere else.) On the Earth's surface the weight of this lunar exploration vehicle is 980 N. On the moon's surface its weight is only 160 N so an astronaut would be able to pick it up and maneuver it by hand. (Acceleration due to gravity on the moon's surface is only 1.62 m.s^-2, or 16.5% of that on the surface of the Earth.) Imagine some component of the lunar lander has a mass of 100 slugs. Its weight on the Earth's surface is 3,220 pound-force (lbf), but its weight on the moon is only 530 pound-force. Some other component of the lunar lander has a mass of 100 pound-mass (lbm). Its weight on the Earth's surface is 3,220 poundals, but its weight on the moon is only 530 poundals. (Acceleration due to gravity at the Earth's surface is about 9.8 m.s^-2 or 32.2 ft.s^-2.) Dolphin (t) 12:41, 1 June 2011 (UTC)[reply]

The component with a mass of 100 pound-mass has a weight of 100 pound-force on the Earth's surface, but on the moon its weight is only 16.5 pound-force. However, these aren't consistent units so applying a mass in lbm and an acceleration if ft.s^-2 in the equation F=ma won't give the answer in pound-force. The answer will be in poundals. Dolphin (t) 06:37, 2 June 2011 (UTC)[reply]

The important thing to remember is that the pound is (historically) a unit of weight (not force or mass). This can be extracted from the Pound (mass) article, but I agree the article could be more explicit on this point. From this definition, we get the pound mass (the mass that has a weight of one pound on the Earth's surface), and the pound force (the force that is equal to a pound (weight)). The slug and poundal come in when we want to avoid putting correction factors related to g in Newton's second law, and keep it as F = ma. Tevildo (talk) 13:01, 1 June 2011 (UTC)[reply]

I found this table helpful:

Three approaches to units of mass and force or weight^[1]^[2]
v t e Base	Force		Weight		Mass
2nd law of motion	$m = .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠F/a⁠$		$F = ⁠ W \cdot a / g ⁠$		$F = m \cdot a$
System	BG	GM	EE	M	AE	CGS	MTS	SI
Acceleration (a)	ft/s²	m/s²	ft/s²	m/s²	ft/s²	Gal	m/s²	m/s²
Mass (m)	slug	hyl	pound-mass	kilogram	pound	gram	tonne	kilogram
Force (F), weight (W)	pound	kilopond	pound-force	kilopond	poundal	dyne	sthène	newton
Pressure (p)	pound per square inch	technical atmosphere	pound-force per square inch	standard atmosphere	poundal per square foot	barye	pieze	pascal

Essentially, pounds can be a unit of either mass or force (or both) depending on the system of units in which one is working. Like the original poster, I was taught (as a physicist) to treat pounds as a unit of force (weight) and use slugs as the corresponding mass unit. This is the FPS-Engineering unit system. It appears in practice, that the pound commonly used to sell products in the United States is usually a measure of mass, however, since scales in stores are checked by placing a reference standard on them and ensuring that the scale gives the correct "weight". Since a standard (of constant mass) is used to calibrate the scale rather than a constant-force standard, this makes the scale essentially a calibrated measurement of mass rather than force. --Srleffler (talk) 18:15, 1 June 2011 (UTC)[reply]

How did people fall into this unfortunate notion that the pound isn't primarily a unit of mass? If someone's having a barbeque on the International Space Station and they ask their buddy to thaw out a pound of hamburger, they're expecting a pound of hamburger, not however many tons it takes to make a pound under their current micro-gravity. If I weigh out a pound of thumbtacks on an old fashioned double pan balance, I'll get the same exact number of thumbtacks whether I'm in Death Valley or Kathmandu. And I dare say most customers expect to get the same number of thumbtacks anywhere in the world or they'd be protesting to the WTO about it. Now yes, I know there are spring balances like you weigh fruit in at the supermarket, but those are so inaccurate that the fact that they're measuring force is just one minor additional detraction. So I don't see where the whole disruptive notion of the pound as a unit of weight force comes from. Wnt (talk) 15:20, 4 June 2011 (UTC)[reply]

Re: "How did people fall into this unfortunate notion that the pound isn't primarily a unit of mass?" For me it was engineering textbooks and university courses. Regardless of how a pound is commonly treated, in engineering it is particularly important to be consistent because if one engineer shares data in pounds with another engineer and the recipient incorrectly assumes he has data with pounds as a force (ie the person sharing treats pounds as mass), the recipient won't multiply by gravity and the building or whatever is being designed may be unsafe by a factor of about 32. Most engineers try to avoid excessive margins of safety (waste of material and cost) so this kind of error may well cause structural failure and death. It is very important that all engineers speak the same lingo (or have a check box on their comps cover sheet specifying whether pounds is a mass or a force). If most engineers are being taught that pounds is a force, then either every engineer must be notified by their governing body that pounds is now to be treated universally as a mass and universities must find textbooks that treat pounds as mass and change their prescribed texts, or people on Wikipedia should refrain from trying to convince inexperienced engineers that pounds is a mass. From the "Foot-pound-second systems of units" section at Pound-force it appears that pound-mass doesn't even play nicely with Newton's second law (F=ma) where in Earth applications "a" is usually directly substituted with acceleration due to Earth gravity (9.81 m/s^2 or 32.17 ft/s^2). Given that F=ma is basically the cornerstone of engineering, if something doesn't agree with it then it shouldn't be used by engineers. I thought I was confused about pounds as a force or mass, but I have since come to realise that I was taught correctly. There is also no reason why someone can't ask for a pound of hamburger with pounds being a force? Obviously if you are on the International Space Station a pound will be different than on Earth, but in space everything is expected to be different. How are go going to tell whether your buddy has stooged you? If you take your trusty kitchen scales (or double-pan balance) with you to the ISS you will get a reading of zero pounds (even if you glue the calibrated mass and the pound of hambuger to their respective pans to stop them floating away). Do the same on the moon and you might beat your buddy up for only giving you 1/6 of a pound. Your scales are only a claibrated measurement of mass on Earth (regardless of measurement in pounds, kilograms or thumbtacks). You should get approximately the same reading in Kathmandu and Death Valley because variation in gravity is too small to be noticeable by most scales (accuracy of your scales may be high but its resolution could still be insufficient), but if you use a precision instrument that displays lots of decimals you will no doubt pick up the difference. Pulling on a spring balance should give you the same reading on Earth and in space, but that is because you are pulling against the stiffness of a spring rather than Earth's gravity, but you still won't be able to use one to weigh your hambuger because it also measures force, not mass (regardless of what the calibrated unit scale on it says). Is there anyone out there who has experience using pound-mass in a practical application where they haven't first converted it to a seemingly more useful system of units (either metric or FPS engineering)? Also, dispite pound being a valid unit of mass in the FPS absolute system of units, how common is the use of a "poundal" as a unit of force? Its not just something that was invented after someone realised the FPS gravitational system wasn't coherent and they weren't willing to concede to pounds being a force? It would appear to me that the FPS gravitational system of units were originally conceived to confuse the crap out of everyone and that Pound-mass has proliferated as an Earth-bound furphy. Would this be a fair assumption or am I just being grumpy now? It would definitely be useful if some of this was cleared up in respective articles. 203.129.23.146 (talk) 02:59, 5 June 2011 (UTC)[reply]

In Niu's "Airframe Structural Design", 2nd Ed (http://www.amazon.com/Airframe-Structural-Design-Information-Structures/dp/9627128090), on page 599 a unit conversion table shows a pound (lb) as a mass of "U.S. Customary Unit" equal to 0.4536 kg. Perhaps the confusion isn't about the system of units, but rather their origin. Perhaps (unverified) a pound is treated as a force in English FPS units and as a mass in US FPS units? This wouldn't surprise me, as the US and UK/English can't agree on another unit of mass, being the Ton. Fortunately they both seem to agree that a ton is a mass, just of different values. 203.129.23.146 (talk) 04:56, 5 June 2011 (UTC)[reply]

In Gere's "Mechanics of Materials", 5th Ed inside the front cover, pound (lb) is tabulated as a force of U.S. Customary Unit equal to 4.44822 N. Even if numerically correct, I guess if the textbooks can't agree whether "pound" is a mass or a force it makes it a bit hard to define. Perhaps all we can do in a Wikipedia article is highlight its ambiguity as either a mass or a force, with conversion factors accurate on Earth's surface. 203.129.23.146 (talk) 05:01, 5 June 2011 (UTC)[reply]

The pound originated at a time when no distinction between mass and weight seemed necessary. Consequently the pound is both a unit of mass (the pound-mass lbm) and a unit of weight (the pound-force lbf). At the Earth's surface an object with a mass of 1 lbm has a weight of 1 lbf. So if someone on the Earth's surface speaks about a pound of potatoes there is no ambiguity about how many potatoes are involved, although to the scientists there is some ambiguity about whether this someone is speaking of mass or weight. Away from the Earth's surface, such as on the International Space Station, a pound-force of potatoes is an extremely large number of potatoes and the on-board scientists are unlikely to have any difficulty understanding what is being said.

These two units are very practical for common usage, but are not consistent units for the purposes of the equations of physics. For consistency in these equations, the lbm and acceleration in ft.s^-2 can be multiplied together to give a weight (force) in poundals. Similarly, the slug (another British mass unit) and acceleration in ft.s^-2 can be multiplied together to give a weight (force) in lbf. Simple. The pound is both a unit of mass and a unit of weight (force). Dolphin (t) 05:27, 5 June 2011 (UTC)[reply]

Thanks Dolphin51. You are making sense to me, and a pound force being equal to a pound mass on the Earth's surface would no doubt be a common cause of confusion. I would also like to correct myself; before I mentioned pulling on a spring-balance on Earth and in space would yield the same result, which would only actually be true for a horizontal pull on Earth's surface, with the person's weight being added in a vertical "hang". Just a technicality. Thanks for the feedback. I've also added a section to the talk page of Pound (mass), where I've tried to highlight the inconsistencies between "pound" as a force/mass even in aeronautical engineering textbooks. I guess I didn't take as much notice during uni, but my lecturers didn't either because I was still taught to use pounds as a force. I'm liking the metric system more and more as I read about the bastardisation of imperial units. 203.129.23.146 (talk) 06:22, 5 June 2011 (UTC)[reply]

Thanks for the positive feedback. I agree that the metric system has a lot in its favor. However, be aware that even though the kilogram is clearly a unit of mass the Europeans have devised the kilopond which is the weight of one kilogram at the Earth's surface! It appears that kilopond is falling out of use to be replaced by the kilogram-force! Dolphin (t) 08:00, 5 June 2011 (UTC)[reply]

No, I don't think so. Among users of the metric system the kilopond/kilogram-force is falling out of use regardless of naming. On the other hand, the FPS-engineering system is still alive, and as its users get exposed to the metric system an analogous engineering variant of the metric system should have obvious attraction to them. I guess this is why the name "kilogram-force", which is an obvious parallel to "pound-force", is getting (relatively) more popular. Hans Adler 14:52, 5 June 2011 (UTC)[reply]

Öveges József, in his popular physics books (such as Kis fizika, 1953), used the name kilogram weight (“kg súly”) instead of kilopond, and the unusual abbreviation “kgs” for it. He also uses units of work, power and pressure derived from this (though also mentions other incompatible units such as horsepower, watts, millimetre of mercury). I don't believe he ever mentions newton (the unit, not the scientist).— Preceding unsigned comment added by B jonas (talk • contribs) 16:12, 5 June 2011

Monoperoxoboric acid and Sodium peroxoborate

What is the structure of monoperoxoboric acid whose formula is [(HO)3B(OOH)] and the structure of sodium peroxoborate whose formula is Na2[B2(O2)2(OH)4].6H2O--Krishnashyam94 (talk) 11:51, 1 June 2011 (UTC)[reply]

I think you might have added one too many hydroxyl groups to the first formula. If corrected, it would be a derivative of boric acid, where one hydrogen is replaced with a hydroxyl group, forming hydroperoxyboronic acid, or monoperoxoboric acid as you call it. The structure of the second formula without the sodium would simply be a dimer of the hydroperoxyboronate anion. The dimer is held together through bridging of two hydroxyl groups in two three-center two-electron bonds. Plasmic Physics (talk) 14:07, 1 June 2011 (UTC)[reply]

Our Sodium perborate article does not agree. DMacks (talk) 15:28, 2 June 2011 (UTC)[reply]

...and now has ref to the xray structure supporting that it's peroxide bridges with normal bonding (no 3c-2e at all). DMacks (talk) 19:33, 2 June 2011 (UTC)[reply]

My mistake, at least it was a plausible theory. Plasmic Physics (talk) 23:37, 2 June 2011 (UTC)[reply]

No worries...would be a reasonable theory. The actual one is an interesting example of how general the ideas of conformation and the cyclohexane-chair are. DMacks (talk) 14:38, 3 June 2011 (UTC)[reply]

Particle accelerator

What is the cost of building a small particle accelerator? --999Zot (talk) 13:54, 1 June 2011 (UTC)[reply]

That's a very open-ended question. Here's a home-built cyclotron constructed as part of a (admittedly very advanced) high school science fair project. In a setup like that, the vacuum pumps will probably run in the hundreds of dollars (purchased used), and likely account for the lion's share of the cost. He reports particle energies up to about 100 keV, which isn't really 'atom-smashing' territory, but is definitely solid proof of principle. From there, the sky's the limit. TenOfAllTrades(talk) 14:26, 1 June 2011 (UTC)[reply]

An historical note: Ernest Lawrence built the first (80 eV) cyclotron for $25, in 1931 dollars (around $350 in modern currency). --Mr.98 (talk) 15:02, 1 June 2011 (UTC)[reply]

Replicating a 1931 design with modern power transistors instead of vacuum tubes should reduce the cost substantially, with adjustments for inflation. Edison (talk) 04:15, 3 June 2011 (UTC)[reply]

Where is the first cyclotron? IIRC, Oak Ridge claims to have it in their museum and the Smithsonian also claims to have it. I've seen both and they look similar. Bubba73 ^{You talkin' to me?} 06:21, 3 June 2011 (UTC)[reply]

How to efficiently get data for alignment from Ensemble?

On this page is an alignment of a gene for multiple animals - I would like each of those genomic sequences, preferably with exons and introns distinct by appropriate capitalisation, in a file that I can then stuff into my alignment program to make a diagram to stick in my report and presentation for people to gawk at. What's the most efficient way? I don't really want to copy and paste individual sequences. --129.215.47.59 (talk) 14:52, 1 June 2011 (UTC)[reply]

What program do you want to put it into? Looking at this I'm thinking that a basic Perl script might do well to process the data, but probably there is a better way. Wnt (talk) 18:09, 1 June 2011 (UTC)[reply]

Lasergene SeqBuilder and MegAlign, but it should be possible to have the sequences in generic formats which most DNA sequence programs can handle? Ensemble doesn't provide a mass export feature for the gene sequences it holds? Isn't it a kind of "must have" feature? --129.215.47.59 (talk) 20:16, 1 June 2011 (UTC)[reply]

Hi! On the left of the ENSEMBL page, you have an option to "export data". Try exporting the alignment to FASTA or another common format, and then you can manipulate it with some sequence alignment viewer, like Bioedit. Hope this helps. PervyPirate (talk) 13:34, 3 June 2011 (UTC)[reply]

Confused about quantum mechanics

I'm confused about quantum mechanics. Postulates of quantum mechanics states that "We can also show that the possible values of the observable A in any state must belong to the spectrum of A." So if the observable for the position on the x axis is x then the "values" of that observable have to be eigenvalues. Let a be one of those eigenvalues, then $x\psi (x,y,z,t)=a\psi (x,y,z,t)$ which is only possible if $\psi$ is identically zero. Now where exactly is my error? 93.132.161.170 (talk) 18:05, 1 June 2011 (UTC)[reply]

In an infinite dimensional vector space such as the Hilbert space of quantum mechanics, spectrum and eigenvalues are not exactly the same thing, although they are closely related. The "eigenfunctions" for the position observable are Dirac delta functions,

\psi (x)=\delta (x-a)

(I have omitted the other dimensions for simplicity). Looie496 (talk) 18:19, 1 June 2011 (UTC)[reply]

Yes, that's it. In my mind I was always restricted to think only of continuous functions. But with the Dirac delta, it is even an eigenvalue/eigenvector. Thanks a a lot. 93.132.161.170 (talk) 19:00, 1 June 2011 (UTC)[reply]

Ouch, I was still confused,

\delta (0)=\infty

, not 1, so this is why it is no eigenvalue and the spectral values come in. 93.132.161.170 (talk) 19:47, 1 June 2011 (UTC)[reply]

Can't help it, the more I think about it the more I get confused again. Could you tell how the Hilbert space and the scalar product are defined in this case? I started trying to understand why a should be a spectral value of x and found I can not do that as long as I don't know "what tricks are allowed", in other words, what the elements of that Hilbert space really are. 93.132.161.170 (talk) 20:28, 1 June 2011 (UTC)[reply]

And now I can cast my confusion into a question again: I understand that $\langle f,g\rangle$ is defined as $\int f(x)\cdot g(x)\,dx$ (for some kind of integral). Now to avoid confusion with the Dirac delta let's denote the Kronecker delta by $\kappa$ . Then $\langle \kappa ,\kappa \rangle =0$ and thus in this Hilbert space, $\kappa$ would be in the same equivalence class as the constant zero function. Now compute $\langle \delta ,f-f(0)\kappa \rangle$ . This would be zero because the right function is zero at point zero. On the other hand, $\langle \delta ,f-f(0)\kappa \rangle =\langle \delta ,f\rangle -f(0)\langle \delta ,\kappa \rangle$ and that would be f(0) as $\langle \delta ,\kappa \rangle =0$ because $\kappa$ is in the same equivalence class as zero. If I don't have an error elsewhere in the computation, I guess I can either use the Dirac delta, picking a point value from a "function" (element of the Hilbert space in question) or I can have a positive definite scalar product defined by integration that ignores point values, but not both at the same time. 93.132.161.170 (talk) 21:05, 1 June 2011 (UTC)[reply]

Are you sure that

\langle \kappa ,\kappa \rangle =0

is correct?

Pretty sure.

0\leq \langle \kappa ,\kappa \rangle =\int _{-\epsilon }^{\epsilon }\kappa ^{2}(x)\,dx\leq 2\epsilon

for arbitrarily small

\epsilon

. If you don't hold with non-standard analysis that only leaves the possibility that the product is zero. To clarify,

\kappa (x):={\begin{cases}1,&x=0\\0,&x\neq 0.\end{cases}}

. Might be the product is defined differently; if so, this is part of my question. 93.132.161.170 (talk) 22:52, 1 June 2011 (UTC)[reply]

I don't see what you are trying to do here. The Kronecker delta, defined thus as a function, is utterly useless. My recommendation is to find a basic graduate-level textbook on quantum mechanics -- this stuff is too intricate to learn from a Wikipedia article or Reference Desk. Looie496 (talk) 23:02, 1 June 2011 (UTC)[reply]

What I did is (if correct) a proof that contradicts the premisses (as I have understood). The Kronecker delta is useful here to construct the function

f-f(0)\kappa

that has point value zero at x=0. So I assume, my premisses (picked from wikipedia) are wrong. 93.132.161.170 (talk) 23:12, 1 June 2011 (UTC)[reply]

The point you are making is correct; the source of the problem is that the way the Dirac delta function is used is physics is not rigorous (the whole concept of the delta functiuon is not rigorous, it has to be defined as a distribution). Then, the position representation should be used with care, in the usual Hilbert space formalism, you can't define it in a 100% rigorous way. You need to work in a rigged Hilbert space to do that.

Now, you can get around these problems by considering position eigenstates |x> to be the limiting cases of physical states that have a finite spread in position. If you formally take the limit at the end of all computations, you never get problems. This is from a physics point of view better because in reality, you can't locate a particle at a precise position. Count Iblis (talk) 23:48, 1 June 2011 (UTC)[reply]

So if I understand you correctly, there is nothing wrong with the postulates but for the usual use of a function space there is trouble having an operator for the position. To get around this, either a more complex Hilbert space is used to handle it strictly correct or (more often, I assume) the Dirac delta is used in the sense of "yes, I know, but all is well because I will take the limit later" just as we multiply and divide by dx (promissing to take the limit later) ? 95.112.146.231 (talk) 00:11, 2 June 2011 (UTC)[reply]

That's right. Typically, you encounter delta functions, not because of working in position representation, but because you work in an infinite volume and then you need to use the delta function normalization. To avoid this, you can put the whole system in a finite volume (which makes momentum discrete instead of continuous), do all the computations with square integrable functions and then take the V to infinity limit at the end. Count Iblis (talk) 01:08, 2 June 2011 (UTC)[reply]

And the Kronecker delta is used to deal with the discrete variables which may come up such as the momentum of a particle in a box example given by Count Iblis above. For such cases the product is defined in terms of a sum - not an integral, and you have

\langle \kappa ,\kappa \rangle =\sum _{i}\kappa (i)^{2}=1\neq 0

. Dauto (talk) 03:11, 2 June 2011 (UTC)[reply]

Is there any textbook on quantum mechanics that explicitly describes what Hilbert space it is using without the reader having to guess from intuition? As seen above, you can virtually "proof" anything if you got the wrong one. 95.112.146.231 (talk) 10:02, 2 June 2011 (UTC)[reply]

Yes, there are such textbooks, typically written for mathematics students. However, you then have to first master the theory of distributions and functional analysis, the prerequisite knowledge is topology and measure theory. I have a book on functional analysis (I don't have it with me right now, I don't remember the title and authors) that has a chapter of quantum mechanics. But, in theoretical physics, we don't teach students by being that rigorous. Students learn quantum mechanics when they have barely mastered linear algebra. And if you want to make progress in theoretical physics, you'd better get used to that, because you won't get the chance to rigorously learn differential geometry before you have to do your General Relativity exam, learn the rigorous mathematical foundations of quantum field theory before it is exam time for that subject.

I therefore think that this book is better for you. I learned QM from that book because I wasn't satisfied with my university textbook. It is considered by many to be the best book on QM ever written. Count Iblis (talk) 15:39, 2 June 2011 (UTC)[reply]

What you say sheds a light on it. So I'm coming in from the "wrong end". I have studied mathematics (but forgot nearly all of it as I didn't have any use or practice for it in the last 20 years). I still have the books (albeit I have to search where I stashed them). Rigorous measure theory is great, especially if you have got sick with "definitions" of probability that go along lines as "tossing a coin very often will finally ...". 95.112.146.231 (talk) 16:55, 2 June 2011 (UTC)[reply]

Rye from Wheat

How can I visually tell rye from wheat? 75.138.213.213 (talk) 22:48, 1 June 2011 (UTC)[reply]

Plant? Seed? Flour? Looie496 (talk) 23:03, 1 June 2011 (UTC)[reply]

Bread? Red Act (talk) 00:09, 2 June 2011 (UTC)[reply]

Beer? 95.112.146.231 (talk) 00:14, 2 June 2011 (UTC) difference-between-wheat-and-rye[reply]

Whisky? SemanticMantis (talk) 14:25, 2 June 2011 (UTC)[reply]

See Rye and Wheat. Assuming you mean the standing crop, rye's awns are longer than those of wheat, but not as long as those of barley. Tevildo (talk) 00:12, 3 June 2011 (UTC)[reply]

^ Comings, E. W. (1940). "English Engineering Units and Their Dimensions". Industrial & Engineering Chemistry. 32 (7): 984–987. doi:10.1021/ie50367a028.
^ Klinkenberg, Adrian (1969). "The American Engineering System of Units and Its Dimensional Constant g_c". Industrial & Engineering Chemistry. 61 (4): 53–59. doi:10.1021/ie50712a010.

[1] Comings, E. W. (1940). "English Engineering Units and Their Dimensions". Industrial & Engineering Chemistry. 32 (7): 984–987. doi:10.1021/ie50367a028.

[2] Klinkenberg, Adrian (1969). "The American Engineering System of Units and Its Dimensional Constant g_c". Industrial & Engineering Chemistry. 61 (4): 53–59. doi:10.1021/ie50712a010.

[1]

[2]