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August 14[edit]

Squiggly rain drops[edit]

I took a picture of the rain outside my window. It was both raining and sunny outside, so I thought that the raindrops would show up in the picture. They did, but they show up as squiggly lines. What is going on here? I took this shot through a window with a basic UV filter on. Camera settings: 1/90s, f5.6, 400 ISO. Any ideas? You will have to zoom into the picture to see the weird effect.

— Preceding unsigned comment added by Dlempa (talk • contribs) 00:31, 14 August 2012 (UTC)[reply]

They're lines because the shutter speed was too slow to "freeze" them. Since the squiggles are all the same shape, the squggliness is presumably the result of camera movement during the exposure. Deor (talk) 00:41, 14 August 2012 (UTC)[reply]

That's actually a pretty cool photo and I 100% agree with the above answer. Vespine (talk) 01:47, 14 August 2012 (UTC)[reply]

You see this effect very often in WWII era dogfight videos, such as here. The video in general may seem only mildly jerky, but the tracer bullets follow seemingly wild, zig-zagging trajectories. Someguy1221 (talk) 01:53, 14 August 2012 (UTC)[reply]

I suggest you use a timer and a stand, to prevent any hand shaking effect. Of course, the camera's own shutter motion might still cause some vibration. You also need a faster shutter speed, but that might make it too dark, so you will need to compensate for that. StuRat (talk) 01:58, 14 August 2012 (UTC)[reply]

I'd expect the circular part of the lines to be due to the shutter movement. An outside force wouldn't be so symmetric or return the camera to it's original position. Interesting to see Ssscienccce (talk) 02:07, 14 August 2012 (UTC)[reply]

But that circular motion occurs in the middle of each rain drop's motion. Wouldn't shutter motion be expected at the start and end, not in the middle ? StuRat (talk) 02:26, 14 August 2012 (UTC)[reply]

Yes, I'm wondering what mechanism could best explain the effect. First idea was focal plane shutters with two moving curtains, the slowing down of the first one and speeding up of the second; but that would show other differences I think. A digital camera can control shutter time electronically without the need for a physical shutter, it could be due to any kind of spring or electromagnet activated movement occuring during the exposure. it just seems too "perfect" to be due to hand movement. But that's just my opinion. Ssscienccce (talk) 04:04, 14 August 2012 (UTC)[reply]

I think that this sort of picture is one in which a slower shutter speed would actually produce a more interesting result (if the camera tremor could be eliminated). If the shutter was so fast as to show dotlike drops, they would be basically indistinguishable from dust specks or other flaws in the photograph. Even to the naked eye, falling raindrops look rather more elongated than the ones in the photo, raising the question of what the "shutter speed" of the human eye is. Deor (talk) 02:38, 14 August 2012 (UTC)[reply]

@ Sturat, the transmission of the shutter shock to the camera's 'retina' moves at the speed of sound in the camera body, while the light moves at the speed of light. It is not inconceivable that the shock would show up in the image after the image was already partially formed. μηδείς (talk) 04:49, 14 August 2012 (UTC)[reply]

Well it would be if the "retina" ie the film reacted instantaneously, but it doesn't. Taking the camera body as an aluminium alloy (speed of sound ~6500 m/s) and the distance from shutter to film as 20 mm, the time taken for shutter reaction to reach the film is of the order 3 microseconds. Time taken for the light, velocity of propagation 300,000,000 m/s, distance 20 mm, the time taken is ~7 picoseconds. Yep, light is darn fast compared to sound. BUT, the exposure time was 1/90 = 11 milliseconds. The nature of film is that the film exposure reaction is sensibly proportional to duration throughout this time. Typically, the shutter will take 1/3 the time to open, and 1/3 the time to close, ie ~3 milliseconds. So, the mechanical delay after the light is about 1/1000th of the shutter opening time, and about 1/3000th of the film exposure time. Clearly the the mechanical delay (and light delay) of of shutter rebound is negligible compared to image formation time - the film has been exposed less than 0.1% by the time the mechanical impulse arrives. The assumptions I've made about distances, shutter opening times etc are very rough, but it is inconceivable that they could be out by a factor of 1000x. Shutter rebound can only affect images by imparting an acceleration to the whole camera (camera sensibly a rigid body) with respect to the scene. Ratbone121.221.30.70 (talk) 05:38, 14 August 2012 (UTC)[reply]

Another part of the effect may be the fact that the sensor in your camera is scanned sequentially. There are some great pictures online of aircraft props taken with iPhones where the prop takes all sorts of interesting forms because it rotated through the lines of pixels as they were scanned. 209.131.76.183 (talk) 11:46, 14 August 2012 (UTC)[reply]

Thanks all for the comments and help! I guess I will continue to experiment. dlempa (talk) 14:53, 14 August 2012 (UTC)[reply]

What is the difference between a "larvae" and a "Nymph"?[edit]

What is the difference between a "larvae" and a "Nymph"? http://en.wikipedia.org/wiki/Special:Contributions/175.140.180.186

See the second and third sentences of Nymph (biology). Larvae, by the way is plural; it's "a larva", not "a larvae". Deor (talk) 02:42, 14 August 2012 (UTC)[reply]

Simply the mode of development. Nymphs undergo incomplete metamorphosis (hemimetabolism), while larvae undergo complete metamorphosis (holometabolism). Both are the juvenile stages of insects between egg and adult.

In nymphs, the growth of the juvenile stages is by gradual acquisition of features of the imago. For example, the first instar may already possess all the features of the adults, except the wings and the ability to produce sperm or eggs. Each successive molt and instar produces larger individuals and the beginnings of wings until the final molt when the imago finally emerges with fully developed wings and full sexual maturity. Note that hemimetabolic insects may also have a stage known as the "prolarva" or "pronymph", which is basically just an advanced embryo. This stage may occur inside the egg, though in some it may actually hatch as prolarva before undergoing their first molt and becoming the first instar nymph. But this time period of a prolarva outside the egg is usually very very short, lasting only a few seconds to a few minutes. The development through nymphs is believed to be the ancestral state in insects.

In larvae, in contrast, the adult features are more or less kept in an embryonic state. They often look like completely different organisms from their adult forms, with very different behavior. They also go through a stage missing in hemimetabolic insects - pupation - where the change from larva to imago is fast-forwarded. It has been hypothesized that this evolved sometime during the early Carboniferous due to the development of eggs with unusually small amounts of food reserves, thus necessitating restriction of growth and requiring the larvae to gather the resources for its final growth on its own. It is also generally accepted (the so-called "pronymph hypothesis") that holometabolic larvae are homologous to hemimetabolic prolarvae (i.e. it's an extended version of the prolarval/pronymphal stage in insects which undergo incomplete metamorphosis).

This is assuming that you're speaking solely about insects. As other animals also use those terms and others differently. Fish young, for example, are also called "larvae", while cephalopod young are called "paralarvae". Ticks, which are arachnids, actually have both larval and nymphal stages, though not in the same sense as in insects as their larvae are just miniature versions of their adults.-- OBSIDIAN†SOUL 03:38, 14 August 2012 (UTC)[reply]

A nymph that lives in a tree is a lot cuter than larvae that live in a tree: [1]. StuRat (talk) 04:30, 14 August 2012 (UTC)[reply]

You do realize that termites do not undergo pupation, and have nymphs, not larvae? Nonetheless, larvae are less attractive than nymphs. μηδείς (talk) 04:44, 14 August 2012 (UTC)[reply]

Higgs field vs. Dark Energy[edit]

What is the relationship between the Higgs field and Dark Energy? Would these balance off against each other to reduce the predicted Cosmological constant to its observed value? Hcobb (talk) 02:47, 14 August 2012 (UTC)[reply]

Both can be explained with scalar fields, although dark energy doesn't require a particle, just a uniform potential as a physical property of spacetime (which is technically much simpler than a scalar field, but similar to the scalar fields used to explain cosmic inflation and the metric expansion of space.) Which predicted cosmological constant are you referring to? 75.166.207.214 (talk) 08:07, 14 August 2012 (UTC)[reply]

I'm attempting to explain the Vacuum catastrophe. Universe starts tiny and empty and then zero-point energy causes Inflation (cosmology) until the Electroweak symmetry breaking limit is reached. At that point a uniform field of Higgs bosons almost pushes through to carpet the universe wall to wall, but is just slightly overmatched by the zero-point energy and the result of the cancellation is the very tiny amount of Dark energy that we see today. Hcobb (talk) 15:31, 14 August 2012 (UTC)[reply]

The Higgs condensate contributes to the zero-point energy, so what you're proposing is really vacuum energy from the Higgs vs. other vacuum energy and cosmological constant. There's an interesting paper from 1980 by Kolb and Wolfram [2] in which they suggest that the vacuum energy of the Higgs condensate could be used to cancel out a cosmological constant; your suggestion goes along very similar lines. However, the famous factor-of-10^120 discrepancy is dominated by other fields, and the Higgs contribution is much too small to make a big difference. It generally makes things slightly worse. Even if you did manage to balance one large contribution against another to get a very small residual vacuum energy, you've only converted the matter into a terrible fine-tuning problem. The real problem is that we don't know how to calculate the vacuum energy even in a well-studied model like the standard one. --Amble (talk) 15:47, 14 August 2012 (UTC)[reply]

Please identify this pill[edit]

One side: G mark or something looking like an on/off button in an almost full circle.

Other side: ML 10 written on two lines.

Size: 5 mm in diameter and 3 mm in width.

Picture link - 1: [URL=http://picturepush.com/public/8971010][IMG]http://www2.picturepush.com/photo/a/8971010/img/8971010.jpg[/IMG][/URL]

[url=http://picturepush.com/public/8971088][img=http://www5.picturepush.com/photo/a/8971088/img/8971088.jpg][/url]

Additional information: The pill was found in my car - After it had been to Hamburg and also after I had been to a festival. It might have been one of my friends that have lost it.

I am pretty sure it's not one of these (Medication my dad is taking) :

zolpidem hexal venlafaxin orifarm furix digoxindak kaleorid tolmin delepsine retard litiumkarbonat oba — Preceding unsigned comment added by Azalin (talk • contribs) 08:58, 14 August 2012 (UTC)[reply]

Only one way to find out. GC-MS it. 203.27.72.5 (talk) 09:03, 14 August 2012 (UTC)[reply]

If you're pretty sure it's something illicit, you could get a kit from DanceSafe or a similar organisation, or alternatively send it to EcstasyData.org or similar for testing. 203.27.72.5 (talk) 09:10, 14 August 2012 (UTC)[reply]

Is it by any chance a hexagonal G ? (It would help to have a picture of that side, too). I found two round, white pills with hexagonal G's alone on one side: donepezil and voriconazole. Here's a 2 page list of those two and 16 other meds with a regular G alone on one side of a white, round pill: [3]. However, none of them matches the "ML 10" on the other side. StuRat (talk) 09:14, 14 August 2012 (UTC)[reply]

That G is the one used by Alphapharm on their tablets, and from their other tablets the 10 will mean 10mg. The product could well be Lumin or mianserin, an antidepressant, which is marked with MI.[4] Graeme Bartlett (talk) 10:15, 14 August 2012 (UTC)[reply]

Sure looks like it: right link this time.. Ssscienccce (talk) 12:22, 14 August 2012 (UTC) oops.. fixed link Ssscienccce (talk) 15:26, 14 August 2012 (UTC)[reply]

Thanks a lot for the help - I believe it's either my Dad who lost it out of his pocket or in worst case - a friend having a depression - I wonder if that pill could be used as a recreational pill. — Preceding unsigned comment added by Azalin (talk • contribs) 23:18, 15 August 2012 (UTC)[reply]

Ternary mixtures and Raoult law[edit]

Considering two binary mixtures, one having negative deviation from Raoult law and the other pozitive deviation, what are the chances that a ternary mixture obtained from the two binaries be almost ideal (no deviation)? --79.119.214.49 (talk) 09:11, 14 August 2012 (UTC)[reply]

It's possible, but it's also possible that there will be a larger deviation in either direction. The deviation is governed by pairwise molecular adhesion and cohesion forces, and you are going from two disjoint pairs ((a+b) and (c+d)) to six pairs (a+b, a+c, a+d, b+c, b+d, c+d). I doubt there is a law that says they will always cancel out in aggregate, but molecular behavior is always a surprise one way or another. 75.166.207.214 (talk) 21:09, 14 August 2012 (UTC)[reply]

All particles moving at the speed of light[edit]

How accurate is it to say that even massive particles move at the speed of light, but because they're constantly bouncing off the higgs field they appear to move slower. Goodbye Galaxy (talk) 14:41, 14 August 2012 (UTC)[reply]

The Higgs field is theorised to give particles mass, and it is the fact that they have mass that prevents them travelling at the speed of light (as massless particles, such as photons, do). I don't know if particles "bouncing off" the field is the right way to think about it, though. I don't really understand the Higgs field, but I've never heard it described that way. The particles are interacting with Higgs bosons, and the end result of that is that they travel slower than light, but I don't think it is because they take an indirect path, which is what you are implying. --Tango (talk) 17:32, 14 August 2012 (UTC)[reply]

I think it's pretty close. The Higgs interaction that gives mass to the fermions is a three-way interaction between a clockwise circular polarized fermion field, a counterclockwise circular polarized fermion field, and the Higgs field. In order for this to preserve angular momentum the two fermion fields have to be propagating in opposite directions. The two fermion fields by themselves are massless. One way of describing this is that a quantum of energy in one of the fermion fields always moves at the speed of light but keeps switching to the other fermion field, reversing direction in the process, at a rate proportional to the strength of the Higgs interaction. (But the real behavior of the particle is a sum over all histories, including all possible times that that interaction could take place. So you shouldn't think of this back-and-forth motion as happening by clockwork, or at random, but rather in a continuous smoothed-out way.)

The mass of the force bosons (W^± and Z⁰) is more complicated, though, and I don't really understand it. -- BenRG (talk) 18:39, 14 August 2012 (UTC)[reply]

It's accurate in a non-rigorous, not-fully-quantum-mechanical picture of how particles behave. A closer analogy is light traveling through water: the light interacts with the water molecules in such a way that it ends up traveling about 1/3 slower than it does through vacuum. The interactions that give mass to a fermion or cause light to travel more slowly through a medium are coherent and continuous. "Bouncing around" could imply discrete scattering events, which would have different effects. --Amble (talk) 19:20, 14 August 2012 (UTC)[reply]

Between this subject and "dark matter" and all manner of other exotic things in physics, I'm waiting for science to decide that there is an Aether after all. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:41, 14 August 2012 (UTC)[reply]

The Higgs field has only a few practical differences from the traditional luminiferous aether involving Lorentz covariance and that it's a medium for fermion inertia instead of photon propagation, in my opinion, although I'm sure saying so is a serious heresy. 75.166.207.214 (talk) 03:33, 15 August 2012 (UTC)[reply]

The Higgs mechanism is not analogous to either light in water or waves in a luminiferous aether. In both of those cases the propagation happens at a fixed speed relative to the rest frame of the medium (water or the aether). Particles that get mass through the Higgs mechanism can move at any sublight speed—all sublight speeds are equivalent in vacuum. It's the particles that don't interact with the Higgs that move at a fixed speed. The nice thing about the back-and-forth explanation of the Higgs mechanism is that it gives a natural reason why you can't exceed c. Everything actually moves at c all the time; you can end up with a slower average speed if you keep changing direction, but you can't end up with a faster average speed. -- BenRG (talk) 05:59, 15 August 2012 (UTC)[reply]

Not exactly: the propagation of light in water is dispersive, and photons can be described as having an "effective mass" in a material. But my main point is that the interaction between the light and water, or a particle and the Higgs field, is not like the random scattering interactions suggested by "bouncing off." Instead, you get a propagating mode that's a coherent combination of the two. That's true for light in water and for the Higgs mechanism. The difference is in the details of the dispersion relation. --Amble (talk) 08:37, 15 August 2012 (UTC)[reply]