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April 29

Beam width of Laser Weapon System (LaWS)

I was just reading about the Laser Weapon System the US Navy is installing on the USS Ponce. I can't find much detail in the Wikipedia Laser articles, but can someone enlighten me as to what sort of beam width is normally used? If you were trying to stop a 6 meter long high speed attack craft, would the laser just slice through it, or melt it, or would it have to hit the engine and explode the fuel? I just want to get an idea of the actual mechanism of a laser destroying a target like that. Thanks in advance. 122.108.189.192 (talk) 07:00, 29 April 2013 (UTC)[reply]

Based on this video of a test firing, it ignites the fuel of the target. 202.155.85.18 (talk) 07:08, 29 April 2013 (UTC)[reply]

...which it could do best with a very narrow beam. Low power InfraRed lasers don't penetrate metal very well - so the power delivered on-target would have to be in the kilowatt range.

In the video of a small drone being shot down by the system[1] - you can see that it makes a small puncture in the fuel tank, then ignites the fuel that pours out.

But they also talk about blinding or dazzling the sensors on target vehicles - which is something that needs a lower-powered, wider beam - so this system probably does some kind of dynamic focussing. SteveBaker (talk) 14:36, 29 April 2013 (UTC)[reply]

The video I link above seems to show 3 separate lasers with one being for heating the target and another for dazzling the sensors. There's no audio and no further explanation to the animation though, so I'm just guessing. 202.155.85.18 (talk) 06:28, 30 April 2013 (UTC)[reply]

This is the OP again. Do you think that means that using the ship-board laser to stop a high-speed attack boat, which would be bouncing over the waves as it approached, would involve trying to hit a vulnerable spot with a narrow beam? (I assume the attacking crew would be hiding behind a reflective screen). That doesn't seem very likely to succeed to me! I just can't understand why there are such high expectations for the system!122.108.189.192 (talk) 07:30, 30 April 2013 (UTC)[reply]

The cool thing about lasers is that you can aim them VERY precisely. All you have to move is a small mirror, maybe just an inch or two across) by some tiny fraction of a degree and you can slew the laser around at almost any speed you want. You can do this with pizoelectrics which can deflect a lightweight object within milliseconds or better. If you use a high-speed camera and off-the-shelf video tracking software on nothing fancier than a regular PC, you can easily know within a small distance where the target is every 1/100th of a second - with military grade stuff, knowing the position to within a fraction of an inch or two every 1/1000th of a second is do-able. The laser itself can be repositioned at that speed too. Since both the detection camera and the laser are interacting with the target at the speed of light, there is no need to lead the target or predict where it's going to be. So stabilising the laser onto the target for long enough to melt a hole in it is not a difficult engineering problem. SteveBaker (talk) 15:44, 30 April 2013 (UTC)[reply]

It sounds like that's the key to it's success, thanks. OP 122.108.189.192 (talk) 07:05, 1 May 2013 (UTC)[reply]

Magnetic field due to rotating sphere

Suppose you have a rotating sphere centered at the origin with given surface charge density $\sigma$ , angular velocity $\omega$ , and radius $R$ . Is there a way to find the magnetic field created a distance $z$ from the origin of the sphere along the axis of rotation?

I have deduced the integral $\int _{0}^{\pi }{\frac {\mu _{0}}{4\pi }}{\frac {2\pi R^{4}\omega \sigma \sin ^{3}\theta }{((z-R\cos \theta )^{2}+(R\sin \theta )^{2})^{\frac {3}{2}}}}d\theta$ , if you consider the sphere as infinitesimally thin loops of current $dI=\omega R\sigma \sin \theta Rd\theta$ , and then the magnetic field due to this loop is $dB={\frac {\mu _{0}}{4\pi }}{\frac {2\pi (R\sin \theta )^{2}dI}{((z-R\cos \theta )^{2}+(R\sin \theta )^{2})^{\frac {3}{2}}}}$ .

a) is this correct? b) is there an easier way? I was wondering if you could use Ampere's law, but there doesn't seem to be a useful choice of curve. AnalysisAlgebra (talk) 10:45, 29 April 2013 (UTC)[reply]

a) Yes it seems to have been correctly derived from the Biot–Savart law.

b) I don't see any easier way either.

c) Do you need any help with the integral?

Dauto (talk) 20:43, 29 April 2013 (UTC)[reply]

Perhaps transforming to the co-rotating frame could simplify the math.Count Iblis (talk) 12:13, 30 April 2013 (UTC)[reply]

The math is not particularly hard, though a bit lengthy. How do you suggest using a co-rotating frame? I don't think I've ever solved a problem of electromagnetism with a co-rotating frame. Dauto (talk) 13:43, 1 May 2013 (UTC)[reply]

Silicic acid

How and when did Jöns Jacob Berzelius discover silicic acid? Plasmic Physics (talk) 11:57, 29 April 2013 (UTC)[reply]

The how is described here which described Berzelius's experiments. Since that source was published in 1836, that would mean the when happened before 1836. This source here states that he was working with silicic acid by 1810, using it to isolate elemental silicon. No definitive answer yet, but some leads. --Jayron32 12:45, 29 April 2013 (UTC)[reply]

This source has a footnote suggesting at least one publication dated from 1839. This may not have been the first time he isolated it, as the two sources above suggested he had already done so, at least by 1836 if not by 1810. But since that source has a specific journal publication by Berzelius himself, you may be able to use that to work backwards to his first isolation of it. --Jayron32 12:52, 29 April 2013 (UTC)[reply]

At first glance, it seems that in first source you listed, he fails to recognise that silicic acid-proper, is distinct from silica, and uses 'silicic acid' to refer to both. Did he not at some later date discover his error, or is that credited to another? This complicates matters, to find when he discovered silicic acid-proper, I must find when he first encountered the 'dissolved silicic acid'. Plasmic Physics (talk) 13:08, 29 April 2013 (UTC)[reply]

The relevant page is 304. Plasmic Physics (talk) 13:09, 29 April 2013 (UTC)[reply]

Berzelius, Jacob (1810). "ZERLEGUNG der Kieselerde durch gewöhnliche chemische Mittel". Annalen der Physik. 36 (9): 89. doi:10.1002/andp.18100360907. and Rose, Heinr (1852). Gedächtnissrede auf Berzelius gehalten in der öffentlichen Sitzung der Akademie der Wiss. Zu Berlin am 3. Jul. 1851. will be a good start to read. (Sorry German speakers only).--Stone (talk) 20:20, 29 April 2013 (UTC)[reply]

Thanks, I'll go through it later today. It's a good thing that I have a rudimentary understanding of written german, I may have to look up a few words. Plasmic Physics (talk) 21:56, 29 April 2013 (UTC)[reply]

Regretably, I don't have access to all articles from Wiley Publishing. Even though the second one is a biography, I'm sure there's something relevant. Plasmic Physics (talk) 23:12, 29 April 2013 (UTC)[reply]

High energy particles

I have read that the universe had an early Radiation-dominated era which has since given way to one dominated by matter. How can we know that there were not ultramassive particles (fundamental or composite) making up most of the mass-energy of the early universe, when they should be larger than any particle accelerator can make? Is there any limit on how massive any as yet unknown particles could be? Is there any sort of equipartition theorem that could apply to how much energy is in rest mass versus relativistic mass in the universe of any era past or present? Wnt (talk) 13:33, 29 April 2013 (UTC)[reply]

You're drawing an arbitrary distinction between radiation and particle. At very high energy scales, the difference is almost non-existent. Very large energy means very large mass, which means very non-localizable "particles..." - particles that behave in a wave-like way. More to the point, thanks to unification, it ceases to even matter when we talk about things like the localization of charge - because, for example, charge interaction and weak interaction are the same above a particular energy-scale. And it ceases to matter when we talk about localization of mass, because we already have a general theory of gravitation that can handle non-localized mass and energy - general relativity. It doesn't matter whether the mass-energy is in the form of a localized "particle" - because we can correctly describe its mass interactions in a more general way. So, the only real missing piece here is the search for a unification between electroweak and strong and gravitational interactions - exactly what is meant by the search for a Grand Unified Theory. Nimur (talk) 16:20, 29 April 2013 (UTC)[reply]

I'm confused. I would think the higher the mass, the smaller the Compton wavelength! Why would they be nonlocalized? And even if nonlocalized, the question of what fraction of c the average particle product should be moving at after being produced from an energy of X should still be valid. Wnt (talk) 16:47, 29 April 2013 (UTC)[reply]

There is a simple rule that applies here. The more massive a particle, the smaller the cross sections for the reactions involving such particles are. To see this, note that in natural units, 1/M has the dimensions of length, therefore cross sections are proportional to 1/M^2. Then what happens is that the more massive a particle is, the weaker the interactions are, which casuses such particles to decouple from the Standard Model particles at earlier times when the universe would be hotter. The sooner they decouple the more energy in total would be in the form of these particles. Since we know how much energy is contained in the dark matter sector, that limits the mass of the dark matter particles. A way out of this argument is to assume that such supermassive particles were never in thermal equilibrium after the Big Bang. Count Iblis (talk) 17:39, 29 April 2013 (UTC)[reply]

But what if they were unstable? For example, I suppose at one point top quarks must have been all the rage, but they presumably have since given way to ordinary protons and neutrons. Speaking of which, can we take from the confident assertions of "lower limits" on the mass of the top quark that these particles can be produced with virtually no kinetic energy relative to their mass, i.e. moving at non-relativistic speeds? Would that have been true in the early universe, so that it might have been "matter-dominated" by vast numbers of top quarks and the like? Wnt (talk) 17:55, 29 April 2013 (UTC)[reply]

The radiation era lasted 10s of thousands of years, while the energy required to create top quarks and heavier particles would have existed for less than a second. Top quarks, and all heavy particles that we are familiar with would decay almost immediately, so they would not be available to contribute to the radiation era. If the origin of dark matter is a WIMP (or something like it) then there would be at least one heavy particle that we don't know about that is stable for the age of the universe. We can set limits on the latter be observing the abundance of dark matter today. Present theories would suggest that any particles heavier than a nucleon are likely to either be very unstable (decaying in less than a second) or very stable (i.e. enduring for the age of the universe). I suppose we can't really rule out the possibility of something unexpected that is stable for thousands of years but not billions, but such an entity would be very unexpected. Dragons flight (talk) 18:21, 29 April 2013 (UTC)[reply]

So are you saying that there was or could have been a matter dominated era before the radiation dominated era, when the unstable quarks were containing most of the mass-energy, but then the universe fell into a gap with (apparently) no stable particles large enough to hold most of the mass at the extant temperature, during which radiation prevailed? (I suppose that by analogy, our own matter based world might gradually spew out more and more energy by solar fusion or evaporating black holes into relativistic neutrinos until most of the mass-energy is in relativistic mass again?) Wnt (talk) 18:46, 29 April 2013 (UTC)[reply]

You might enjoy reading timeline of the big bang, there were many qualitatively different phases, though several of them were compressed into the first second or so. I don't think there could have been a "matter-dominated" era prior to the photon period, in part because "radiation-dominated" in this context means not only photons but also any particles moving at relativistic speeds. I'm pretty sure that the majority of matter present in the very earliest periods would have been relativistic. The transition between the "radiation-dominated" epoch and the "matter-dominated" epoch isn't actually about a change of type exactly. As the universe expanded after the Big Bang, the density of normal matter decreases as

1/a^{3}

where a is the scale factor of the universe. In other words, a larger universe with the same amount of matter leads to decreasing matter density. However, photons and other relativistic matter also suffer a gravitational redshift as the universe expands around them. As a result, the mass density of photons goes as

1/a^{4}

rather than

1/a^{3}

. It is this different scaling law that causes the density of radiation to fall below the density of matter, and hence transition to a matter dominated universe. While it is certainly true that stars partially convert the mass energy of their atoms into radiation, I don't think the effect is nearly large enough to make a difference in the over all balance. More likely we will move (or have already moved) into an era that is dark energy dominated rather than matter dominated. Dragons flight (talk) 22:58, 29 April 2013 (UTC)[reply]

That's a good point. Do isolated relativistic particles also slow down/lose energy as spacetime expands? Yet ... I'm thinking that if you look back far enough (or have a Big Crunch in the future) the higher-energy particles will collide, and in colliding, produce still higher-energy particles in some kind of equilibrium. What's weird about our time is that we have matter, not antimatter, and the matter takes an absurdly long time to decay if at all, so it's out of equilibrium. But is this the only such time, the only such mechanism? Wnt (talk) 03:03, 30 April 2013 (UTC)[reply]

Muscle mass

Would one lose a significant amount of muscle mass and fitness if they stopped gym training for 3 weeks? Clover345 (talk) 15:11, 29 April 2013 (UTC)[reply]

Muscle mass, no. Older people lose muscle mass faster than young people, but even for the elderly there shouldn't be much loss in three weeks. Fitness, it depends on the details. A person who lies in bed for 3 weeks may lose substantial fitness, but with ordinary levels of everyday activity there shouldn't be much of an effect during that time. Looie496 (talk) 15:25, 29 April 2013 (UTC)[reply]

Like Looie says, this is almost impossible to quantify without specifics, and even then difficult to predict. There are, however a lot of folk beliefs among athletes/weightlifters about this kind of thing that some quick googling will reveal. I tried looking on pubmed for studies related to this but I'm having a hard time thinking of how exactly this would be described... "cessation", "inactivity"... can't find anything yet. From personal experience I would say yes, you would notice a loss of strength after 3 weeks off, although it wouldn't take that long to get back to where you were either, depending on your level of fitness. Shadowjams (talk) 18:49, 29 April 2013 (UTC)[reply]

Yes, also from my own personal experience, you would see a lot of changes, even after a few days of not exercising if you exercise close to your physiological limit. Fitness would actually increase after several days due to super-compensation effects in the absense of the usual heavy workout load. You'll notice that your resting heart rate drops to levels that are much lower than your typical resting heart rate. But after two weeks or so, your fitness levels will start to go down, although you will not notice this as you are not exercising. Only when you start to exercise will you notice such effects. Suppose that you use to run for half an hour at a fast pace. Then what will happen after 3 weeks is that more than 15 minutes into the run it all get suddenly lot more difficult. Normally, if you had a decline in fitness while regularly exercising (e.g. due to a cold), you would not see such effects. You would then automatically run at the right pace and you would more gradually feel that it is time to stop.

This is why people who exercise a lot are adviced to keep on exercising even when they have a mild illness like a cold, but then at a lower intensisty and for a shorter time. Athletes who break their legg will often do swimming to stay fit. Count Iblis (talk) 12:08, 30 April 2013 (UTC)[reply]

Auroral oval

Why do auroras on the Earth (or any planet for that matter) form in rings? In particular, why do they not form directly at the magnetic poles? I understand the basics of the auroral mechanism, in that charged particles from the solar wind get caught in the magnetic field and are directed towards the poles, then interacting with the Earth's atmosphere, but why do they form *around* the poles as opposed to *at* the poles?

-- — Trevor K. — 18:22, 29 April 2013 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

Aurorae form along magnetic fields. This image illustrates that the Earth's field is not concentrated at the poles at the surface (that is, the Earth is not an ideal bar magnet 8000 miles in length). — Lomn 18:43, 29 April 2013 (UTC)[reply]

The Earth's magnetic field spreads before it reaches the surface.

It is useful to consider the shape of the Van Allen radiation belts along the lines of the Earth's magnetic field. Auroras aren't actually extensions of those (now three!) belts but are somehow based on connections between the solar flux lines with other magnetic field lines. So the particles, coming from "outside" the area of the Van Allen belts, end up coming into the poles further north/south than the spot where those belts would meet the atmosphere (if they did). Wnt (talk) 18:54, 29 April 2013 (UTC)[reply]

(ec) Earth's magnetic field has an approximately cylindrical symmetry, well modeled as a dipole magnet with a cosine-law and an offset angle. Earth's atmosphere has an approximately spherical symmetry. The aurora occur when trapped particles, traveling along field lines, interact with neutral gas in the atmosphere, colliding and luminescing. The intersection of the field-lines with an isobaric altitude of the atmosphere - in other words, an L-shell surface intersecting with a spherical shell - approximates an ellipse (this can be shown using analytic geometry, or just by looking at a diagram); that's the reason you see auroral rings. Nimur (talk) 18:56, 29 April 2013 (UTC)[reply]

Auros actually appear near closed/open magnetic field line boundary(so called, polar cusps) . It is about 60° for the Earth's field and a typical interplanetary magnetic field. This boundary is a special place because it is related to magnetic reconnection, electron acceleration and transport of magnetic flux from the dayside to the nightside part of the magnetosphere and back. Ruslik_Zero 02:35, 1 May 2013 (UTC)[reply]

Smart OR beautiful women

Here in Russia we have a popular meme stating: "there are only two types of women: 1) smart, 2) beautiful. There are almost no women with both characteristics".

It's a sexist statement, and I do believe it's wrong (I personally know a lot of exceptions from this "rule").

I would like to know how popular is this meme in the World. The patriarchic Russia only?

Has this statement been scientifically proven as wrong? (a positive correlation between IQ and some beauty score or something) --Roman1969 (talk) 21:23, 29 April 2013 (UTC)[reply]

The better question is whether it's been proven right... which is unlikely. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:22, 29 April 2013 (UTC)[reply]

Well, to prove it wrong, one needs only to find one woman who is both smart and beautiful. Categorical generalizations are easy to disprove, especially when the terms are so subjective! SemanticMantis (talk) 22:52, 29 April 2013 (UTC)[reply]

Not exactly, as the quote says "almost none", which is a weasel term. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:18, 30 April 2013 (UTC)[reply]

This would suggest that in fact attractive women (and men for that matter) are more intelligent. Mikenorton (talk) 23:13, 29 April 2013 (UTC)[reply]

I find that the more intelligent women are more attractive. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:18, 30 April 2013 (UTC)[reply]

In English speaking countries we talk about "dumb blondes" and share jokes about dumb blondes - this is really the same nonsense as the Russian meme. I've never thought it remotely true. In my experience, the prettiest women are usually the more intelligent. However, I think the reason put forward in in article cited by Mikenorton is not the whole picture. The more intelligent girls a) are better able to select clothes that suit their facial type and body shape, and b) they work in jobs and with work colleagues that encourage them to take more interest in their apperance. Choosing the right hairstyle to suit her face shape and clothes that suit a girl's figure can make a big difference to how attractive she looks. The more intelligent girls work in better jobs and can spend more money on makeup and clothes too. A fantastic personality can overide a plain appearance.

An exception to what I've said does obviously occur. A small number of girls go on beauty courses. A girlfriend I had once worked as a hotel receptioniste in a large city hotel. The hotel manager selected "front of house" employees on the basis of their attactiveness, but you need no great brains to do the work, just a pleasant manner, confidence, and the ability to cope with a smile with obnoxious shitty guests. My girlfriend was the prettiest girl I've ever met, but she really was a dumb blonde who claimed her large makeup costs as a tax deduction.

Wickwack 124.182.168.245 (talk) 01:14, 30 April 2013 (UTC)[reply]

If she was successful, it doesn't sound like she was that dumb.... Nil Einne (talk) 01:46, 30 April 2013 (UTC)[reply]

You might ought to read A Spell for Chameleon. μηδείς (talk) 02:42, 30 April 2013 (UTC)[reply]

Danica McKellar anyone? --Jayron32 05:29, 30 April 2013 (UTC)[reply]

There may be other factors here. A woman who works hard so that her male colleagues know she is "not dumb", will waste less time making sure her hair looks perfect, etc. etc. In my own family the women who don't work are the ones who will typically say that those who do don't dress properly, that their hair looks bad etc. Also, women (and men) who are scientists may on average have less interest in cosmetics. In a BBC horizon documentary about a decade ago, it was suggested that women in technical jobs behave socially more like men while men in jobs like nursing behave socially more like women. A scientist who wakes up in the morning may be far more inclined to think about her work and not even think about looking in the mirror to see if she looks ok, while someone who works in a factory, who does work that is not so interesting, will look in the mirror and even be slightly late for work because of problems with her hair or makeup. Count Iblis (talk) 11:46, 30 April 2013 (UTC)[reply]

That does not match my experience. I've had the privilege of working with a number of women in professional engineering roles, and have met several women medical specialists working at professor level. They wear little or no makeup at work, and they dress conservatively, but they still take a lot of care with - and have pride in - their appearance - hair styles, clothes, etc. It is just not at all necessary for a woman to work hard just to show men she's bright, any more than a man needs to. If you are bright, it will be evident in conversation, and in the quality of your work, regardless of whether you are male or female. It does, however, appear a little difficult for women to interact like men, as they think differently - one of the reasons for the so called "glass ceiling" restricting women from senior managment. Ratbone 124.178.140.70 (talk) 12:28, 30 April 2013 (UTC)[reply]

Here's a recent study on the relationship, suggesting that the positive correlation that I mentioned above is stronger for men, but still clear for women. They talk about other research that links both these attributes to underlying genetic fitness. Mikenorton (talk) 12:41, 30 April 2013 (UTC)[reply]

I married a Russian-speaking blonde. She was certainly no dummy; she got rid of me, if any proof of her intelligence is called for. -- Jack of Oz ^[Talk] 21:12, 2 May 2013 (UTC)[reply]

Can anyone identify this flower?

What is its common name and scientific name? I found it at longwood gardens Thanks so much

— Preceding unsigned comment added by 153.104.177.176 (talk) 22:26, 29 April 2013 (UTC)[reply]

Looks like an Azalea, but there are thousands of cultivars, so it will be very hard to track it down further. See many similar images by doing a google image search for /orange red azalea/, like so: [2] SemanticMantis (talk) 22:49, 29 April 2013 (UTC)[reply]

Longwood Gardens. μηδείς (talk) 02:41, 30 April 2013 (UTC)[reply]

Looks like Clivia miniata to me. Oda Mari (talk) 09:21, 30 April 2013 (UTC)[reply]

I can't find an image of an azalea with a yellow throat but the Clivia miniata images do match perfectly. Richard Avery (talk) 06:48, 1 May 2013 (UTC).[reply]