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July 31

Visibility of the New Moon

The Wikipedia entry for New Moon states 'The Moon is not normally visible at this time except when it is seen in silhouette during a solar eclipse'. While driving from California to New York on Interstate 10 I pulled over early in the morning in the desert to stretch my legs. I look up at a perfectly clear night sky and saw the entire surface of the moon un-illuminated. What phase of the moon was I viewing? — Preceding unsigned comment added by 66.65.43.46 (talk) 00:56, 31 July 2013 (UTC)[reply]

If the Moon is closer than 7.5 degrees from the Sun, there is no visible crescent see here. Count Iblis (talk) 01:11, 31 July 2013 (UTC)[reply]

If we assume that your recollection is accurate, then you saw a lunar eclipse—at night you can't see the new moon because the Earth is in the way. TenOfAllTrades(talk) 01:32, 31 July 2013 (UTC)[reply]

Yep, that's the only possibility. In addition to what you've quoted regarding the new moon, it's also only above the horizon during daylight hours (layman's definition, not astronomical). It's impossible to see a new moon, or see a dark disc where a new moon would be, at night. — Lomn 01:38, 31 July 2013 (UTC)[reply]

No! No! A new moon can be above the horizon shortly after dusk or before dawn, and it is illuminated by earthlight, which is easily bright enough to make it visible in good conditions. (Earthlight is over ten times as bright as moonlight.) Looie496 (talk) 02:32, 31 July 2013 (UTC)[reply]

In an area with high levels of nighttime illumination (such as in large cities), it's also possible for a new moon to be darker than the night sky, and therefore visible as a dark circle. 24.23.196.85 (talk) 02:54, 31 July 2013 (UTC)[reply]

Skyglow over cities is caused by light that originates on Earth, and is then scattered back down from within the Earth's atmosphere. The Moon cannot obstruct this light, because the light goes nowhere near the Moon. So, no dark spot in your city sky. TenOfAllTrades(talk) 03:26, 31 July 2013 (UTC)[reply]

It is possible but unlikely that he saw it during a total lunar eclipse. But then it would have been reddish, and he didn't report that. Bubba73 ^{You talkin' to me?} 02:36, 31 July 2013 (UTC)[reply]

I have seen a full lunar eclipse (as well as a full solar eclipse) and it was not a eclipse. I was in the high New Mexico desert at the time, late November perhaps 4 or 5 AM, and it seemed to me that it was being illuminated by the Earth's albedo. I have posed the question because there seems to be differing views as to what I saw is in fact possible. — Preceding unsigned comment added by 66.65.43.46 (talk) 02:54, 31 July 2013 (UTC)[reply]

That's not a bad hypothesis, and it's one I've thought about too. Unfortunately, the sky at twilight – after the Sun has set, but before the sky is fully dark – is bright enough to completely wash out the Earth-lit new Moon. The Moon can't get much further than about 7 degrees from the Sun before it starts to show a visible crescent, which means that even under ideal conditions you're just at the edge of civil twilight. (Go ahead and search—you'll find that there just aren't any images of a new, non-crescent moon at twilight. I will also note, for the record, that I have another reason to hate Stephenie Meyers—her silly vampire books and movies hopelessly contaminate Google image searches for new moon and twilight.) TenOfAllTrades(talk) 03:34, 31 July 2013 (UTC)[reply]

A quick question or two: (1) it was in the early morning - was it just before sunrise? (2) Was the Moon in the same direction as the rising Sun? A new Moon would be close to where the Sun would rise. In a lunar eclipse, the Moon would be opposite the Sun. Bubba73 ^{You talkin' to me?} 04:01, 31 July 2013 (UTC)[reply]

And what was the date on which you saw this?--Shantavira|^{feed me} 07:34, 31 July 2013 (UTC)[reply]

List of lunar eclipses here. In 2012 there was a penumbral on nov 28, mainly visible between 14:00 and 15:00 UT, that would be between 7:00 and 8:00 local time? Ssscienccce (talk) 09:50, 31 July 2013 (UTC)[reply]

I don't believe that one can see a perfectly unsunlit moon for the reasons given above, but it's certainly possible to see a very thin crescent moon with the rest of its visible disc faintly illuminated by earthshine. This has traditionally been referred to as "seeing the new moon in the old moon's arms", as in the ballad of Sir Patrick Spens. There's a good image here. Deor (talk) 12:46, 31 July 2013 (UTC)[reply]

To somehow detect Everett's "other worlds": some new "telescope?"

Hi, I'm certainly a layman, but i was thinking that if Hugh Everett's Many worlds interpretation does turn out to be true, then physics needs a new "telescope", that is, an instrument as radical as the telescope(and microscope too) were in the 1600s, in order to "see", that is, somehow detect the other worlds. I'm wondering if the new "telescopes" could be experiments at temperatures much colder than they've done yet. I mean, I have heard of physicists getting down to .1 degree kelvin, but maybe the clues could be there at .00001 kelvin, or even 10^(-40) kelvin? And what are the hopes right now for getting these much lower temps? Thanks76.218.104.120 (talk) 09:29, 31 July 2013 (UTC)[reply]

I just read a reference to a proposed experiment to detect them by Plaga! He would use ions in some way.76.218.104.120 (talk) 09:35, 31 July 2013 (UTC)[reply]

Please forgive me for not checking, wikipedia says they are down to a few hundred nanokelvins now. I got up late at night with an idea that was exciting to me and i didn't read up. But the general direction of my first and second questions still stand, but they are much closer to 10^-10 than I thought. So, my second question can be revised to what would be the outlook for even lower temps, for example 10^-10000 k? Thanks again.76.218.104.120 (talk) 09:50, 31 July 2013 (UTC)[reply]

The problem is that the different interpretations of quantum mechanics yields the same predictions for the outcome fo experiments. The only way to test the Many Worlds Interpretation (MWI) is to think of some very unusual experimental set-up that would yield a different prediction. As long as you have the usual type of quantum experiments where the observer is treated classically, you will not see any difference. The only way to get a difference is to think of an experiment where the observer is itself going to participate in a non-trivial way in the epxeriment where a de-facto calssical description of the observer (due to quantum decoherence) is not a valid assumption anymore.

David Deutsch came up with such a thought experiment back in 1985. This involves an observer who can be perfectly isolated from the environment. He prepares an electron with its spin polarized in the x-direction. He then measures the z-component of the spin. This measurement can yield two outcomes with equal probability, spin up or spin down. In the MWI both measurement outcomes really exist, but you'll find yourself in one of them. So, how would you know that the other outcome also exists? Well, since you are perfectly isolated, you could in principle reverse all the steps made in the measurement process, ending up with the electron spin polarized in the x-direction. However, you would not know anymore that you had previously measured the spin of the electron, as your memories would also have to be reset in that reversal process.

What you can do is keep in your memory is the information that you did perform the measurement, the transformation back to the initial state is then still a unitary transform (an easy exercise to check) and thus relaizable in principle. You cannot keep the result of the measurement; because this is different in the two branches where the measurement result is different, you won't get the electron back in its original spin state polarized in the x-direction. This verifies the physical existence of the two different measurement outcomes, because both are needed to get back to the initial state.

Of course, this experiment as originally proposed by Deutsch is impossible to perform in practice, however a few years later people came up with the concept of a quantum computer. If very large scale quantum computing ever becomes a reality, you can contemplate simulating an artificially intelligent entity inside a quantum computer and that entity could then perform Deutsch' experiment. Count Iblis (talk) 13:45, 31 July 2013 (UTC)[reply]

thanks, very good exposition.76.218.104.120 (talk) 21:51, 1 August 2013 (UTC)[reply]

Is the moon in the sky exactly as often during the day as during the night?

That is, if I timed how long the moon was in the sky during the day, and how long it was in the sky during the night, over a long enough period, would I get identical answers?

Be as picky as you want about details, and do please try to quantify your answers: I'm interested in whether the various skewed orbits and other factors make a difference. 86.164.26.17 (talk) 10:02, 31 July 2013 (UTC)[reply]

You should to a first approximation. Now, you see more of it at night because the full moon is at it's zenith at local midnight, while the new moon is at zenith at local noon. But averaged out over an arbitrarily long time, there are no stretches of any 24 hour cycle when the moon is in "the sky" more than at other times. --Jayron32 11:16, 31 July 2013 (UTC)[reply]

You never get things like that exactly the same if you measure closely enough. Lets just consider the moon to be a point to simplify things. The sun isn't a point so that makes days longer than nights. Refraction makes the light go further so that also makes days longer. Discounting those two factors the earth has a width so with a point sun the day is a little bit shorter than the night if it rotates evenly. The earth goes in an ellipse round the sun which I believe means the northern hemisphere gets slightly more daylight as the earth is farther away from the sun and goes slower when it is tilted towards the sun. Discounting the tilt as well we have the problem of the speed of the moon as it orbits the earth, I think it should be going slightly faster when it is on the sun side so that would mean it is more often out at night. And I haven't started on earthquakes or the other planets. Anyway the most important factor discounting refraction is the width of the sun I believe so you would have it more often in the day than the night. Dmcq (talk) 13:27, 31 July 2013 (UTC)[reply]

Seems that Dmcq has already given the reasons for the two not being equal: sunrise and sunset are defined as the moment the upper edge of the sun is at the horizon, and atmospheric refraction. When considering twilight, the difference becomes larger, see definitions here. I calculated the total time the moon would be visible at 0°W 20°N during 2013, using the tables you can generate here; results: moon visible during the day for 134937 minutes (2248.95 hours or 93.706 days), and during the night for 128363 minutes (2139.38 hours or 89.14 days). Some other numbers: total daytime 184.68 days, nighttime 180.31 days; moon for 182.85 days, no moon for 182.15 days Ssscienccce (talk) 13:40, 31 July 2013 (UTC)[reply]

Same calculation for 0°W 60°S gave: moon during the day: 132561 minutes, during the night: 125251 minutes. (92.06 vs 86.98 days) Total daytime 185.18, nighttime 179.82 days; moon for 179.04 days, no moon 185.96 days total.

Of course the start and end date will influence the results since the moon's position won't be the same after 1 year..Ssscienccce (talk) 13:58, 31 July 2013 (UTC)[reply]

The moon is always in the sky, isn't it? Mingmingla (talk) 16:46, 31 July 2013 (UTC)[reply]

Are you counting the new moon? Because although the moon is in the sky during the day, a human eye will not be able to see it. Also the thin crescent moons are not that obvious during the day, so are you counting that as being in the sky, as you have to look carefully to notice it. So the moon is certainly less noticeable in the day than at night. guess you knew that! Graeme Bartlett (talk) 21:51, 31 July 2013 (UTC)[reply]

Over a long period of time (probably on the order of some years) the moon will average the same period of time over any longitude during day as night. That's intuitively obvious, but you can ask for an explicit explanation at the math desk. μηδείς (talk) 18:46, 1 August 2013 (UTC)[reply]

Electron Affinity of flourine< chlorine, why?

Why is the electron affinity of flourine low then that at chlorine? — Preceding unsigned comment added by 202.70.84.89 (talk) 15:17, 31 July 2013 (UTC)[reply]

If you're talking about the Pauling electronegativity, then fluorine does not have a lower electronegativity than chlorine, but a higher. Plasmic Physics (talk) 22:41, 31 July 2013 (UTC)[reply]

No. That's a related, but different parameter. See Electron affinity. Dominus Vobisdu (talk) 23:12, 31 July 2013 (UTC)[reply]

My apologies, I haven't thought critically about these things for a long time. I have heard of electron affinity, the term just escaped my mind prior your link. Plasmic Physics (talk) 00:45, 1 August 2013 (UTC)[reply]

The reasons why fluorine is anomalous are explained in detail here: [[1]]. Dominus Vobisdu (talk) 23:12, 31 July 2013 (UTC)[reply]

The easiest explanation, although it may not be completely correct, as I was taught to memorize and recite, is that the 2p orbital you would have to insert the electron into experiences significantly more electrostatic repulsion with the electron that is already in that orbital. While the effective nuclear charge for chlorine's 3p orbital is lower than for fluorine's 2p (which would explain why fluorine has the greater electronegativity), it has much less electrostatic repulsion than fluorine's 2p orbital, which makes the chlorine 3p orbital the lower energy state.--Jasper Deng (talk) 00:53, 1 August 2013 (UTC)[reply]

F itself isn't the special-case/anomaly. F being low (or Cl being high) in F→Cl→Br→I is the same pattern as O→S→Se→Te (O is low or S is high) and likewise for the carbon and boron groups. The link Dominus Vobisdu mentioned mentions this pattern for the second row vs third row. DMacks (talk) 01:55, 1 August 2013 (UTC)[reply]

Utilization of infrared, solar cells.

Greetings!

I am somewhat puzzled by the advent of solar-power generators capable of producing electricity from the infrared part of the spectrum. For starters, I've heard that they can generate power in all kinds of weather—even at night! Also, I've heard that the only reason we (end users) don't have them is because they produce electricity at about 10 THz, making them totally useless with which to run our appliances, which typically run at approximately 50 to 60 Hz.

I have two quick questions; namely, how exactly can a solar cell generate power at night? Does it utilize infrared light, from the Sun, absorbed by Earth's atmosphere during the daylight hours, or perhaps, infrared radiation arriving here from more-distant stars? Also, (and I apologize in advance, if this is stupid question) just how impractical and unwieldy would it be to just nix all of 50-60 Hz electronics and replace them with 10 THz ones—the way we did with regular gasoline, and are currently doing with analog television?

Would be the cost greatly outweigh the benefit of living free of power plants, dams, coal mines, transmission lines, et al? Thank you.Pine (talk) 20:57, 31 July 2013 (UTC)[reply]

Sounds unlikely. Do you have a source? Maybe it got a bit distorted in transmission or in recall. The frequency you say it produces electricity is only a little lower than for infrared, and certainly it could not be transmitted over power lines. See Terahertz radiation, Infrared. Photovoltaic Solar cells produce direct current. Mirror arrays can concentrate sunlight on a boiler and probably could produce power from infrared if there were enough of it hitting the mirrors, but residual heat in the atmosphere, or moonlight, or starlight would not be enough. And if the mirror array did boil water, then the generator would produce plain old electricity unless you somehow had a microwave generator or like which ran on heat, or the concentrated black electromagnetic energy just heated something up so it radiated black body radiation. Edison (talk) 21:34, 31 July 2013 (UTC)[reply]

I found an article Third generation photovoltaic cell which mentions advanced solar cells making electricity from infrared at night, but it links to a dead reference, so it is less than convincing. Edison (talk) 21:40, 31 July 2013 (UTC)[reply]

Solar cells can respond to near infrared with wavelengths shorter than about 1 micron. You don't get much of this at night at all, as it comes from the sun like normal visible light. When you get the longer wavelength thermal infrared, solar cells do not turn this into electricity in any significant amount. THey are just as likely to do the reverse process and radiate thermal infrared. The issue here is entropy. Note there are other techniques such as solar thermal power. Graeme Bartlett (talk) 21:42, 31 July 2013 (UTC)[reply]

I found an IEEE article from 2012 which discusses Terahertz power generation. It generally has less than 1% conversion efficiency in generating the Terahertz output, and such EM radiation is reduced severely as it travels through the atmosphere. Edison (talk) 22:05, 31 July 2013 (UTC)[reply]

The OP may be referring to so-called nano-antennas that receive the IR energy as electric signals which after rectifying would result in usable energy. Very promising it seems "Unlike solar cells, nano-antennas can operate round the clock, independently of weather conditions such as humidity and cloud cover and without restriction of orientation towards the sun.", but when the rectifier is discussed, the optimism wanes: "The implementation of rectennas for energy harvesting at IR frequencies has remained an elusive research area due to the limitations of nanoscale fabrication and inability to implement rectifiers that could handle EM radiations oscillating at trillion times a second." And I assume the solar panels would have to be raised in the air (and receive IR from the ground), otherwise the laws of thermodynamics would prevent "round the clock" operation. Ssscienccce (talk) 00:52, 1 August 2013 (UTC)[reply]

The downward IR radiation for a visibly clear sky is about 70% of the upward IR radiation from the ground. Dragons flight (talk) 06:38, 1 August 2013 (UTC)[reply]

Yes, that's the reason why panels on the ground can't generate motive power at night; it would violate Carnot's theorem. Ssscienccce (talk) 19:53, 1 August 2013 (UTC)[reply]

How do you determine if a bird is fat?

I was just looking at this photo and it made me think. How do you determine the difference between a really fat bird and a bird that's just very fluffy? Is there a reliable thing to look for? --87.114.229.248 (talk) 22:06, 31 July 2013 (UTC)--87.114.229.248 (talk) 22:06, 31 July 2013 (UTC)[reply]

Several methods are possible, of which the two most obvious quick and sloppy methods are weight/length ratio and thigh circumference/length ratio. These presupose that you have reference data for the specific population in question. If you mean by mere sight, an experienced observer could probably make an educated guess. Again, depends on long-term observation of the population. You can find more sophisticated measurement methods like ultrasound and x-rays by Googling measuring body fat content in birds. Dominus Vobisdu (talk) 22:36, 31 July 2013 (UTC)[reply]

You could kill it and use standard food analysis techniques. Although bird lovers may not like that approach, IRWolfie- (talk) 23:52, 31 July 2013 (UTC)[reply]

I have chickens who confuse me the same way. Three are basically identical, except there's a small, a medium and a large. Now and then, the small one looks bigger than the big one. While they're ruffled, it's hard to tell by sight. I'd say feel it, but that's obviously not easy with an owl.

In this guy's case, note how deep the branch is into his body. Probably safe to assume that's not flab, and that he's at least equally feathery on the other side. I'm no expert on fish owls, but he doesn't look particularly fat to me. InedibleHulk (talk) 00:39, 1 August 2013 (UTC)[reply]

Can he fly uphill or only downhill? --catslash (talk) 00:42, 1 August 2013 (UTC)[reply]

Can birds get fat at all? Birds have a very high power to weight ratio, allowing them to fly. So, it seems to me that if they could get obese, that would compromize their ability to fly. Count Iblis (talk) 01:05, 1 August 2013 (UTC)[reply]

If you've ever had the pleasure to cook a goose, you would know that some birds, at least, can accumulate an enormous amount of fat. Dominus Vobisdu (talk) 02:09, 1 August 2013 (UTC)[reply]

Birds that fly can't get all that fat, no. The leaves us with flightless birds and chicks, where fat may be advantageous for the usual reasons (insulation, surviving periods of famine, etc.). StuRat (talk) 06:07, 1 August 2013 (UTC)[reply]

I think the OP is using "fat" colloquially, as in the opposite of thin, not necessarily the opposite of skinny. There's a great clip on youtube from a japanese tv show, you can find it if you search for transformer owl. it shows an owl trying to make it self look "fat" and then trying to make it self look skinny, it's quite incredible. Vespine (talk) 01:42, 1 August 2013 (UTC)[reply]

I don't know if it is inferable, but I once had a cockatiel which would get exhausted from just circling three times within our twin garage. We took it for a fly several times a week, just like you would take a dog for a walk. Maybe it was just not very fit when it came to flying, but not fat? Plasmic Physics (talk) 01:45, 1 August 2013 (UTC)[reply]

Of course you can get truly fat birds: ducks for example are prized for their fat and along with geese are "fattened" on purpose for fois gras. I imagine penguins too would have some fat for insulation. But I agree those would be the exceptons amongst wild birds, expecially if you only count flying ones. Vespine (talk) 06:28, 1 August 2013 (UTC)[reply]

Here is an interesting cross-section showing just how much of an owl is feathers: [www.neatorama.com/2012/04/17/owls-are-all-feathers/‎] 209.131.76.183 (talk) 11:40, 1 August 2013 (UTC)[reply]

I am sorry, but that has to be about the ugliest non-moulting non-chick I have ever seen. And apparently the inspiration for Big Bird. 18:43, 1 August 2013 (UTC)

I wouldn't call it ugly, but it won't win a beauty contest either, I'll resign to calling it odd-looking. It brings to mind what the result could be if you visually combine a housecat with a hawk. Plasmic Physics (talk) 01:11, 2 August 2013 (UTC)[reply]

I second all of that. My first thought was that it had the exact same eyes as one of my housecats (who is hairy and unquestionably obese, but not quite repulsive). InedibleHulk (talk) 06:22, 4 August 2013 (UTC)[reply]

More like a Swifter and a used Hoover bag with a beak. Or Phyllis Diller and a yellow Peep. μηδείς (talk) 01:16, 2 August 2013 (UTC)[reply]

The Chicken Lady is more what I think of if Diller got any birdier. No murdered women in this one, but you may still find her disturbing. Discretion advised. InedibleHulk (talk) 06:28, 4 August 2013 (UTC)[reply]

A more serious answer is yes, you can tell if a captive bird is in good condition by catching it, and feeling for how "full" the big muscles either side of the breast bone are. I'm not sure how to find an illustration, it may be just something you learn by experience. Any bird breeder can probably show you how to do this. Snake experts use the same sort of guide by looking at how full the muscles are beside the backbone, and a rating for "fatness" of Koalas has been developed by feeling either side of their spine in the lumbar region. 122.108.189.192 (talk) 07:57, 3 August 2013 (UTC)[reply]