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December 31

Black Holes!

What would happen if you had a space ship out in intergalactic space, and they tied a rope around an astronaut and dropped him thousands of light-years into a black hole (the space ship is out of danger etc etc, and the astronaut doesn't starve). He falls past the event horizon.

Does the rope continue to play out? How far does it play out? Does it go 1 event-horizon radius and then go slack (or at least not be pulled anymore by the astronaut.. to say nothing of gravitational effects on the rope itself), or is it "bigger inside"?
How much force is put on the rope? Would it pull the space ship toward the black hole?
If the astronaut, after falling past the event horizon, pulled on the rope with all his strength, would his crewmates playing out the rope feel the tug? Could he communicate in morse code to them or something?

--⁪ffroth 01:56, 31 December 2007 (UTC)[reply]

The ship's crew would see the astronaut slow down (and become more and more redshifted) so that he approaches the event horizon but never crosses it. From their point of view, neither the astronaut nor the rope ever touches the event horizon, so the experiment gives them no new information about the inside of the black hole (as required by the definition of "event horizon").

The experience of the astronaut himself depends on what the ship's crew eventually does. If we assume they can't keep holding the rope forever, then they must either pull him back up, or let go of the rope (or the rope breaks somehow, which amounts to the same thing). If they pull him back up, then nothing particularly unusual happens (except that the astronaut will have aged less than the crew). If they let go of the rope, then the astronaut falls in and gets to see the inside of the black hole, but he can't tell the crew about his experience, not even by tugging Morse code on the rope, because it's no longer connected (P.S. also because the crew has already died of old age).

The one important point is that unless the ship's crew allows the rope to continue paying out forever (by their watches), the astronaut never gets across the event horizon. In any event, the crew never observes anything cross the event horizon. That's why the word horizon is so appropriate, because you can never observe anything cross it. —Keenan Pepper 04:34, 31 December 2007 (UTC)[reply]

So how did "stuff" get inside the event horizon in the first place in order to create the black hole? I've always failed to grasp the theory of black hole creation by gravitational collapse for this very reason. Spinningspark (talk) 17:21, 1 January 2008 (UTC)[reply]

The original star that became the black hole had "stuff" - otherwise it wouldn't have stayed over the Chandrasekhar limit (I'm not sure I got that right) and become a black hole. Effectively the event horizon means that things get inside, but nobody sees them get inside because their time moves much, much, much faster than ours. I remember the thought experiments about "what if the moon fell into a black hole, astronauts included", and apparently the astronauts would get to see the universe age with their own eyes at superspeed as they fell in, but after that, who would know ('cause it's a singularity). Altho there is a very small chance that would happen. ~user:orngjce223 how am I typing? 19:48, 1 January 2008 (UTC)[reply]

Yes, the original star had "stuff". But from our inertial frame it will take forever to collapse beyond its own event horizon. In its own inertial frame (or an astronaut falling in) it happens in a finite normal time, but to an outside observer it does not. So given that a black hole exists and postulating that it was formed by the collapse of a super heavy star, it must have happened forever ago in our frame. Which leads to the conclusion that black holes existed before the creation of the universe. This is a contradiction, or if it is not, no ione has ever succeeded in satisfactorily explaining it to me. Spinningspark (talk) 23:42, 1 January 2008 (UTC)[reply]

So, you could just pull someone/something out from beyond the event horizon of a black hole using a rope? --Kurt Shaped Box (talk) 22:43, 1 January 2008 (UTC)[reply]

I don't want to put words in anyones mouth but I think the previous posters were saying just the opposite. Spinningspark (talk) 23:45, 1 January 2008 (UTC)[reply]

Yeah, I think I misread the response slightly. Still, how could they continue playing out the rope, yet not have the astronaut in question eventually cross the EH? Surely it's just a question of falling a certain distance in a certain period of time? I mean, does the astronaut actually fall any slower in reality, just because the effects of the black hole are (to the observer) screwing with human time perception? EDIT: Yes, I realize that I may be moving the discussion along to a complex debate on the existence of reality outside of human perception.... ;) --Kurt Shaped Box (talk) 01:52, 2 January 2008 (UTC)[reply]

No, the spaceship NEVER observes the astronaut cross the EH. The descent of the astronaut becomes slower and slower from their perspective as the EH is approached. This is down to time dilation, the astronaut also appears to age very slowly. The astronaut, from his perspective, on the other hand, is falling at a perfectly normal rate but the spaceship and everone on it is ageing very fast. This scenario is similar to the twins paradox only gravity is the cause of the time dilation in the case of black holes. Spinningspark (talk) 07:14, 2 January 2008 (UTC)[reply]

The postulate of distorted space time at the event horizon still requires proof. I postulate that if the spaceship observed the astronaut freeze at the event horizon, then the spaceship would also observe other matter at the event horizon. Nearby stars would reflect light off that matter and thus a black hole would not appear to be black at all. If a black hole 'lived' - lets say almost as long as our Universe, it would also seem fair to expect enough matter would have been pulled in for the event horizon of such a black hole to be observed as an asteroid belt - orbiting nothing. Often, movie special effects show matter being stretched as it is pulled in and beyond the event horizon, but in fact, the spaceship cannot see light beyond the event horizon (even if once inside the horizon - the astronaut is dematerialised). Assuming that an event horizon is clearly demarcated (and that the effects are not gradual over a long distance), the astronaut would approach and hit the event horizon - then disappear (rapidly fade out - as some light photons will reflect towards the event horizon). If the astronaut had (some sci-fi here) an anti-gravity spacesuit and was able to survive inside the event horizon, then try to look out beyond it, I would imagine all he would see is a blur of light as particles of light are pulled in from all directions, stretched, scattered and warped. The astronaut would no longer be able to see his spaceship as a clear object. -- LordVonPS3 12:40, 12 August 2009 (UTC)[reply]

Careful, there's been some misinformation in this thread. I'm not sure whether the astronaut is being slowly lowered or whether he's bungee-jumping. I'll assume bungee-jumping. In that case, as others have said, the ship sees the moments just before the astronaut crosses the event horizon stretched out and redshifted into the indefinite future. (Classically, anyway. In reality the astronaut emits only finitely many photons before falling through, and the last one will reach the ship after a finite, and typically very short, time.) This is the Doppler effect, not time dilation. The astronaut does not see a reciprocal blueshift. He sees only a part of the ship's history before falling in, and more of its history after. (You can see out of a black hole from inside.) The ship does not have an unlimited time to pull the astronaut back out. The time in the ship's history that the astronaut sees when crossing the event horizon is also the moment at which it becomes impossible for the ship to do anything to save the astronaut. After that, although they still see light from before the astronaut falls through, it's causally impossible for them to save him; in effect he has "already fallen through". I hope these ugly ASCII-art diagrams will make things clearer.

This is a portion of a Kruskal-Szekeres diagram of a black hole (or something like that, anyway). The diagonal line is the event horizon (a null line). The vertical line is the astronaut (this needn't be straight, it's just hard to draw otherwise). The dotted line is the ship; its worldline is a hyperbola with the event horizon as one asymptote. Light from the astronaut to the ship travels like this:

            |      /.
            |     /.
            |    /.
            |   /.
            |  //.
            | //.
            |///.
            ///.
           /|//.
          / |/.
         /  |/.
        /   |/.
       /    |.
      /     |.
     /      |.
    /       .

It's hard to see in this picture, but the hyperbolic shape of the ship's worldline means that light emitted arbitrarily close to the event horizon reaches the ship arbitrarily far in the future (unbounded redshift) and light from behind the event horizon never reaches the ship at all. Light from the ship to the astronaut [error corrected —KP] travels like this:

            |      /.
            |\    /.
            |\\  /.
            |\\\/.
            |\\/\.
            |\/\.
            |/\\.
            /\\.
           /|\\.
          / |\*
         /  |\.
        /   |\.
       /    |.
      /     |.
     /      |.
    /       .

The * is the point in the ship's history that the astronaut sees when crossing the event horizon, and also the first point at which the ship can no longer take action to save him (since the effect of any action is limited to its future light cone). In practice the point of no return occurs far earlier, since real materials only have so much tensile strength and the force you need to save the astronaut goes to infinity as you approach the critical time.

The hole will continue to pull the rope in indefinitely; it's just like throwing anything else into a black hole. Suppose the ship decides to stop releasing the rope some time after it's too late to save the astronaut. Then, assuming they can stop the rope at all, it will exert a continuous pull on the ship for as long as they continue to hold it. This pull has a familiar name: weight. The weight of the rope is ∫ a dm = ∫ ρ(r) a(r) dr. The acceleration a goes to infinity as r approaches the event horizon, but the rope does not extend that far; it has necessarily broken somewhere above the horizon where its tensile strength was exceeded. Since the rope is broken, the ship can't use it to signal the astronaut, but it can signal him by other means (like a light beam). The astronaut can't signal the ship by any means. -- BenRG (talk) 09:11, 2 January 2008 (UTC)[reply]

Reason for shift in apparent solar midday at winter solstice

The sun begins to set later about 10 days before winter solstice, and the sun continues to rise later until about 10 days after solstice. In effect, this shifts the apparent solar midday later around the time of the winter solstice. Can anyone explain to me, in layman's terms, why this happens? (I have read the article Equation of time and largely failed to understand it.) Thanks. Marco polo (talk) 02:42, 31 December 2007 (UTC)[reply]

What you have to understand is that the Sun's movement in the sky that you see every day is not only caused by the Earth's rotation. Most of it is, but a small fraction is caused by the Earth moving in its orbit around the Sun. Think of a diagram of the Earth in its orbit. In one day, the Earth has moved 1/365 of the way around its orbit, or a little less than 1°, right? But that means that, over course of a day, the Sun is now in a different direction, as seen from the Earth, than it was. It's moved by about 1°. So the Earth has to rotate through almost 361°, not 360°, to bring the Sun back to the same place in the sky. (The difference between the two amounts adds up to exactly 360° per year, corresponding to the Earth making one revolution around the Sun. The time to rotate 360° is called a sidereal day and there is one more of those in a year than the ordinary or "solar" day.)

Okay, now the tricky part is that the extra amount that I called "about 1°" is not the same every day. This is because when the Earth orbits around the Sun, it does not move in an exact circle (the distance to the Sun changes by about 3,000,000 miles from nearest to farthest) and it does not move at a constant speed. So on a certain date the Earth might have to rotate by only 361.1° (say) to bring the Sun to the same place in the sky. That means that instead of 24 hours from one solar midday to the next, it takes 24 x 361.1/361 hours, and the solar midday shifts later by 24/3610 hours or about 24 seconds. On another date at another time of year, the Earth only has to rotate 360.9° from one solar midday to the next, and midday shifts back the other way against the clock. I just used 0.1° and 24 seconds as an example; I don't know what the actual maximum of the daily shift is.

These midday shifts are going on all year (except for the times when the shift happens to be zero), and the cumulative shift can get to about 15 minutes either side of the "middle". But you only notice it near the solstice because it's when the length of the day is nearly constant that you see the sunrise and sunset shifting the same way.

--Anonymous, 05:50 UTC, December 31, 2007.

Thank you: I understood that! Marco polo (talk) 15:53, 31 December 2007 (UTC)[reply]

If you were to take a picture from the same place at 12:00 noon every day (ignoring Daylight Saving Time/Summer Time), the pattern of the sun's locations would be called an analemma. That article may help give a visual interpretation of what's happening. -- Coneslayer (talk) 16:55, 31 December 2007 (UTC)[reply]

Galactic Rotation Curves influenced by time dilation?

Hi, I was just reading an [article] about how giant voids in inter-galactic space possibly have time running faster than spaces that house galactic clusters. Because more mass concentrated in space means times runs slower in that space. So, looking at Galactic_rotation_curve, I'm wondering maybe the central part of the galaxy has time running slower than at the outer fringes. In that case, the velocity of the inner stars would be slower, right? In that case, maybe there is no Dark Matter needed to explain how the rotation velocity appears constant over the galactic radius. Just a thought... what do you think? --InverseSubstance (talk) 05:44, 31 December 2007 (UTC)[reply]

Yes, but that effect apparently does not account for the entire effect, now, does it? Otherwise they wouldn't have designed big elaborate apparata (is that the plural of apparatus?) to house various experiments to detect possible dark-matter candidates. ~user:orngjce223 how am I typing? 19:53, 1 January 2008 (UTC)[reply]

Since you ask, apparatus is (iirc) a u-stem noun (like manus, sinus, impetus, flatus, status, fetus, cantus) and its nominative plural is apparatūs. (Neuter u-stem nouns have ū, ua. The more common o-stem nouns have us, ī or, if neuter, um, a.) —Tamfang (talk) 00:52, 7 January 2008 (UTC)[reply]

Magma affected by gravity?

Is the molten rock inside the earth affected by the pull of the moon? Does this cause tides in there? What are the effects of these? Do they create earthquakes, or siesmic shifts? Harland1 (^t/_c) 08:05, 31 December 2007 (UTC)[reply]

Sure. See earth tide. Because these are spread over a large area, they don't directly "cause" earthquakes. The earthquake article explains the causes of earthquakes. The amount of rock that is molten is relatively tiny, by the way.--Shantavira|^{feed me} 09:39, 31 December 2007 (UTC)[reply]

Thanks, yeah I do know what causes earthquakes I was merely wondering whether his could be another cause. by the way I had always thought that most of the mantle was molten, presumably this is wrong then? Harland1 (^t/_c) 12:06, 31 December 2007 (UTC)[reply]

The mantle is nearly all a deformable solid with properties analogous to cheese. It deforms and even flows under external stresses, but left alone it is firm and holds its shape. I'm not sure where the myth of a "magma ocean" comes from, but it is a very pervasive misunderstanding. In actuality, only a tiny portion of the mantle is molten (i.e. liquid) and nearly all of that is at hotspots near the surface. Dragons flight (talk) 12:19, 31 December 2007 (UTC)[reply]

Thanks tut tut tut.. I learnt all magma was molten at school! Harland1 (^t/_c) 13:09, 31 December 2007 (UTC)[reply]

All magma is molten, but most of the Earth's interior is not magma. --Sean 14:58, 31 December 2007 (UTC)[reply]

Something with the consistancy of cheese would indeed be pulled and pushed around by the tides - but think about it: If this was the cause of earthquakes, we'd see earthquakes that occurred on a regular approximately twice-daily basis just like the ocean tides. Since we don't see earthquakes that happen with that degree of regularity - it's pretty safe to assume that tides in the solid earth are not related to earthquake activity. SteveBaker (talk) 16:45, 31 December 2007 (UTC)[reply]

labelled figure of laminar air flow -biotechnology

59.95.121.219 (talk) 12:44, 31 December 2007 (UTC)[reply]

What exactly is your question? SteveBaker (talk) 16:39, 31 December 2007 (UTC)[reply]

This is from our boundary layer article. Hopefully it helps. Nimur (talk) 03:25, 1 January 2008 (UTC)[reply]

Could I walk on the Sun, hypothetically speaking

Could I walk on Sun, i.e is there a ground of the sun that could support any solid object, if it could resist the heat??

Thank you in advance. I couldn't read that out from the Sun article —Preceding unsigned comment added by 82.131.61.107 (talk) 14:02, 31 December 2007 (UTC)[reply]

The sun is a ball of gas and plasma as far as we know. Depending on the density of the gas/plasma, you may be able to float or swim in it; but I dont think walking would be possible (Not to mention the gravitational field at its surface (does it have a surface?))--TreeSmiler (talk) 14:29, 31 December 2007 (UTC)[reply]

The sun#Structure mentions that it is not solid but a plasma.--TreeSmiler (talk) 14:33, 31 December 2007 (UTC)[reply]

The sun's "photosphere" - which is what we see as the "surface" of the sun (although it's not solid) - is at a temperature of around 6,000 degrees C. The highest melting point substance we know of is (I believe) diamond - which melts at 4,440 degrees C if you place it under enough pressure. At 'normal' pressures, Tungsten survives the highest temperatures at 3,422 degrees C. So, no, there is nothing that would resist enough heat to make it into the suns photosphere as a solid. You could conceivably use some kind of ablative material that's designed to boil away and thus carry off the suns heat for a while - that would allow you to visit the photosphere briefly. In the center of the sun, the temperature gets up to over a million degrees - so that's right out of the question. However, humans or present day electronic robots would not survive at the photosphere even if they could be kept cool enough because the amount of radiation of all kinds coming from the sun would be fatal in a very short time. I suppose something with very thick lead shielding encased in a large amount of ablative diamond might survive long enough to be interesting - but whatever instruments you carried to examine the sun would be cooked by the heat and radiation long before they could get any interesting readings - so this wouldn't be much of a useful mission! SteveBaker (talk) 16:37, 31 December 2007 (UTC)[reply]

Read the novel Sundiver for a fairly realistic description of a journey to the Sun's chromosphere. —Preceding unsigned comment added by Keenan Pepper (talk • contribs) 19:29, 31 December 2007 (UTC)[reply]

I think the OP was asking a hypothetical question which amounts to "is the surface of the sun dense enough to hold up a human?", not how quickly would he fry if he tried it. According to the sun article, the average relative density is 1.4. Since a human body is mostly water with a density of 1.0, I conclude that a human would float on the surface. You would not get in many breast strokes though, in the microsecond before you were destroyed by UV radiation, pressure, solar flares etc, etc. Spinningspark (talk) 17:39, 1 January 2008 (UTC)[reply]

This oversimplifies it somewhat, as the sun will become denser the closer you are to the centre, due to gravity. If you could (somehow) resist the temperatures, pressures, ionising radiation etc. involved then I'd imagine that you would fall down into the sun (with your velocity undergoing an exponential decay due to the gradually increasing upthrust) until you got to the point where the density of that plasma surrounding you was equal to the average desity of your body (or whatever vehicle or container you were in). If any one has some data about the sun and understands all the physics involved (I'm not sure I do, but I might give this a shot anyway) then it might be an interesting exercise to calculate how far the average human would fall into the sun before slowing down to a stop and 'floating'. JMatopos (talk) 15:56, 6 January 2008 (UTC)[reply]

Paleonthologists

I'm looking for the full names of the following paleonthologists, which I haven't been able to find either in Wikipedia or in Internet.

L. Ginsburg
K. Oenigswald
S. Torch
L. Ister

Thanks in advance. -- Danilot (talk) 15:34, 31 December 2007 (UTC)[reply]

I can't promise I found the ones you're thinking of, but all of the following published paleontological papers (oh the wonders of Google Scholar):

Léonard Ginsburg
(You may have spelled his name wrong) Wighart von Koenigswald. There was also a GH. R. von Koenigswald from way back (I found citations to papers in the 1930s, but I'm not sure what his full name was).
(Again, possible typo here) I couldn't find any S. Torch, but there is a Petr Storch
Couldn't find any L. Ister, but there are plenty of Listers. Someguy1221 (talk) 16:30, 31 December 2007 (UTC)[reply]

Thank you a lot. Those are probably typos as the same source cited Heinz Tobien as T. Obien. -- Danilot (talk) 17:28, 31 December 2007 (UTC)[reply]

Note also that it's paleontology', no "H". --Sean 23:24, 31 December 2007 (UTC)[reply]

Wow. – b_jonas 14:59, 1 January 2008 (UTC)[reply]

Below 5% Alcoholic Beverage

Do beverage with below 5% alcohol have effects?whats their measures?Flakture (talk) 16:04, 31 December 2007 (UTC)[reply]

What exactly do you mean by "effects"? And what do you mean by "their measures"? Might just be me, but I don't really get the question. Anyway, yes, they certainly have an intoxicating effect, although that effect is of course smaller than beverages with a higher percentage of alcohol. Aeluwas (talk) 16:17, 31 December 2007 (UTC)[reply]

Well, there is still alcohol in drinks labelled "below 5%" - if you drink enough of them, you could still get drunk or end up over the legal limit for driving. The question is: How far below 5% ? SteveBaker (talk) 16:19, 31 December 2007 (UTC)[reply]

Beer is often below 5% alcohol so, yes, it clearly has effects. Perhaps you were thinking of Low-alcohol beer which has less than 0.5% alcohol? 16:22, 31 December 2007 (UTC) —Preceding unsigned comment added by Rmhermen (talk • contribs)

That would make more sense - but the answer is still the same. The drink is labelled "less than X% alcohol" because it's not zero. Hence, if you drink enough of it, you'll still be over the legal limits for driving (for example). But it's hard to say how much unless one knows the exact percentage. Certainly there are beers down below 5% and if your "low alcohol" beer is actually at 0.5% then it would take ten bottles of the low alcohol stuff to equal one bottle of the full strength brew - which means that you'd have to drink an outrageous amount of the stuff (maybe 20 bottles!) to get over the drink-driving limit (depending on your local laws of course) - and you'd probably be physically unable to drink enough to get yourself actually drunk. But if you really are talking about "less than 5%" then it's a different matter. SteveBaker (talk) 18:11, 31 December 2007 (UTC)[reply]

Well there is a great one called Carling C2 in the UK. That's only 2% but does a convincing job of tasting like their full-strength variant. Of course as the above say all drinks with some % of alcohol will have an effect, but you would need to know the specific drink and the laws of your country to figure out how much you can have. For more info on measurements look at Alcohol (proof) ny156uk (talk) 20:06, 31 December 2007 (UTC)[reply]

Presumably that redlink should be Proof (alcohol) AndrewWTaylor (talk) 20:28, 1 January 2008 (UTC)[reply]

What would the surface of the sun look like?

In science fiction, they generally show the surface of a star looking like hot lava. I use to think it was a gas, which doesn't look anything like that, but apparently it's plasma. Would it look something like that? — Daniel 17:15, 31 December 2007 (UTC)[reply]

We do have actual pictures of the sun, like this one. Commons actually has a ton of them, see commons:Category:Sun. Someguy1221 (talk) 17:26, 31 December 2007 (UTC)[reply]

Plasma is essentially just gas where the electrons have been stripped from the atoms - it looks (as you might expect) like brightly glowing gas - kinda similar to a flame (but flames are not hot enough to be plasma). You see plasma more often than you might think: lightning, the aurora borealis, the sun, plasma-screen TV's, plasma cutters and welders, those glass 'lightning discharge' balls...all over the place! Check out plasma for some pictures. But as for the surface of the sun, it's hard to say how something that you cannot safely look at would look. In order to see anything at all without burning your retinas, you have to have processed the incoming light in some way. In that case, what you see depends entirely on how the light was processed - you can never know what it "really" looks like. SteveBaker (talk) 18:03, 31 December 2007 (UTC)[reply]

The sun doesn't really have a surface, as such. It's a plasma (as has been mentioned - if you don't know what it is, imagine a glowing gas or something) that just becomes more and more dense the closer you get to the centre. If you were just wanting to know what the photosphere of the sun would look like to the "unaided" eye, then it would just be a extremely bright white light. Your eyes would stop working entirely before you got a decent look at it though. JMatopos (talk) 16:04, 6 January 2008 (UTC)[reply]

Why does it appear to have a surface, rather than a continuous density gradient? —Tamfang (talk) 01:06, 7 January 2008 (UTC)[reply]

As far as I know, because at a certain range of temperatures (and hence, at certain distances from the centre) the plasma emits white visible light that we see. This is the part normally called the photosphere. JMatopos (talk) 13:09, 7 January 2008 (UTC)[reply]

There's more info on the Photosphere page, if you want it. JMatopos (talk) 13:40, 7 January 2008 (UTC)[reply]

Airplane safety (2)

I've posted a question about airplane safety on Dec 23. And many of the replies sugest that a cushion or heat insulator would be to expensive to be used on general plane. Do you know of aerogel? It has a very good insulator property and it's one of the lightest material in existence. I also know that it could held a 2.5 kg bricks just using 2 grams of aerogel. So I think it's a pretty good property to be used for cushion, what do you think? roscoe_x (talk) 18:15, 31 December 2007 (UTC)[reply]

It's also somewhat absurdly expensive, if I recall correctly. I'd actually inquired about insulating my house with it, before being informed just how much that would cost me...(I don't quite remember the figure, it was a while ago) Someguy1221 (talk) 18:33, 31 December 2007 (UTC)[reply]

I don't know how much it is now - but a couple of years ago when I enquired, a cube of the stuff 3" on a side cost $500. United Nuclear are charging $5 for 100cc of irregular shaped granules and say they cannot find anyone who will sell it to them in blocks. However, it's very badly suited to airplane safety - firstly it's made of glass and it shatters just like glass does - secondly it's extremely hygroscopic - it absorbs liquids like a super-sponge due to capilliary action through the tiny voids that make up it's structure. Hence in a crash, if it didn't shatter into dust, it would immediately soak up all of the fuel and mix it very efficiently with the air in it's own structure - which would probably make the most efficient fuel/air bomb imaginable. This would be a total disaster in an airplane crash in which a fuel tank ruptured! So, no - for every possible reason - aerogel would be a disasterous material to use for thermal insulationn in a plane. SteveBaker (talk) 19:06, 31 December 2007 (UTC)[reply]

Kalwall claims to sell the stuff commercially as thermal insulation [1]. I don't know what it costs, but it must be lower than the prices cited above or no one would buy it.

Atlant (talk) 03:42, 1 January 2008 (UTC)[reply]

From memory multiple other flaws in the suggestion were pointed out Nil Einne (talk) 08:15, 1 January 2008 (UTC)[reply]

Do genital integrity groups absolutely oppose religious circumcision?

Since the subject is so contentious, it's hard for me to understand the real viewpoint from the various WP pages. My main question is, theoretically, would GI groups be picketing every brit milah? Or do they give Jewish rituals a bye, reasoning that Jews don't promote circumcision among Gentiles? Do GI folk view every Jewish newborn male as a victim to be saved? Rpresser (talk) 19:14, 31 December 2007 (UTC)[reply]

I don't think GI folk are monolithic; they probably have different opinions, depending on their respect for religion. Some might say that religion is a crock and not something to waste time and money "mutilating" babies over. I, however, couldn't possibly comment. --Seans Potato Business 20:31, 31 December 2007 (UTC)[reply]

I don't know about "groups" in general - I certainly oppose it. I don't care what the bullsit religious grounds are - mutilating a child's body before they have even a chance to make up their own mind over it is sadistic and ought to be flat out illegal. It's hard to say whether my opinion is typical of those who oppose such things - but I suspect it is. SteveBaker (talk) 00:33, 1 January 2008 (UTC)[reply]

SteveBaker: Would you also oppose circumcision of boys in sub-Saharan Africa - where (male) circumcision has a proven positive effect in reducing HIV infection? --Psud (talk) 02:21, 1 January 2008 (UTC)[reply]

Is the point to protect the child himself, or to create herd immunity? If the former, it seems reasonably safe to say that the child will be old enough to make up its own mind before catching HIV from intercourse becomes a major risk. If the latter, it seems pretty harsh and draconian, really. Not to be crude, but personally, I would not be interested in having my dick cut up for the common good. APL (talk) 18:50, 2 January 2008 (UTC)[reply]

While I can't say for sure my impression is most people from the group will agree with SB. If anything religious (and cultural) grounds is one of the worst grounds. HIV reduction for example or other genuine health grounds would probably be the only sort of reason why people from such groups may say it was acceptable (although many would probably disagree with the evidence or merits of such a public health approach). Note that the practice is not limited to Jewish people it is also common among Muslims. And I don't see how religious practice would matter since the newborn baby obviously didn't choose to be born into whatever religion he was born in to. While opinions of religions must vary significantly most people would probably still stay harmful religious practices should be stopped and for these people clearly the practice is harmful. I don't know if it's accurate to say they would see 'Jewish newborn male as a victim to be saved'. Sure those intrinsicly opposed to religion may see that. But there are probably many who simply want to end what they see as a harmful religious practice even if they see religion as inherently good. In other words they want to see Jewish people and Muslims carrying out this practice stop doing so, rather then 'save Jewish and Muslim newborn males' Nil Einne (talk) 08:23, 1 January 2008 (UTC)[reply]

I think it's safe to assume that if a group considers an action immoral, it will not make exceptions based on religion only. Religion has simply been the excuse of too many crimes. Would the U.S. Department of Homeland Security, for instance, excuse the 9/11 attacks because they were religiously motivated? It's also hard to imagine even a multicultural country, for example Canada, allowing honor killings because religious immigrants arrived. A crime is a crime; that's what most people believe. --Bowlhover (talk) 00:35, 3 January 2008 (UTC)[reply]

Many would consider circumcision a victimless crime, however. Victimless not in that there is no target for the act, but in that the obvious victim doesn't care. Let me explain; I am not referring to the obvious fact that the baby doesn't complain much. This is a bit anecdotal on my part, but I have never known a Jew to appear traumatized or resentful toward his parents for having a mostly useless piece of skin removed. This is much unlike the female counterpart to circumcision. Someguy1221 (talk) 03:27, 3 January 2008 (UTC)[reply]

I know many circumcised men who wish they had their foreskin still. 64.236.121.129 (talk) 18:55, 3 January 2008 (UTC)[reply]

How about David Reimer for the traumatic? Then again, things like that don't happen very often209.151.139.22 (talk) 05:47, 5 January 2008 (UTC).[reply]

Need Ocean Animal Info

I am doing a report for school and having trouble finding names of animals that live in the Hadalpaligic (Deepest) zone of the Ocean. Any help would be appreciated. Thanks in advance. —Preceding unsigned comment added by 68.55.154.229 (talk) 19:14, 31 December 2007 (UTC)[reply]

Well, spelling it correctly would be a start. It's called the hadopelagic or hadal zone. —Keenan Pepper 19:31, 31 December 2007 (UTC)[reply]

confuesed

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

We really can't offer you a diagnosis or recommend (or discourage) particular treatments. You should take any concerns you have to a medical professional whom you trust, like your school nurse or your family doctor. TenOfAllTrades(talk) 20:43, 31 December 2007 (UTC)[reply]

Passage number

Passage number has been defined as the number of times that a culture has been subcultured. Technically, this should be wrong, right? I have a master culture, and take 100 aliquots from it, send them around the world for friends and family to set up their own subculture, then my master culture has been subcultured 100 times. I know that the definition is trying to infer the number of subculturings that went into making the subculture in question but that's still not what it says. The definition is only understandable if presented in the correct context, and with some "fuzzy processing". I tried to come up with a better definition and failed (poor vocabulary). I would like someone here to define it (for Wiktionary). --Seans Potato Business 20:28, 31 December 2007 (UTC)[reply]

All you have to do is define the drawing of aliquots (and subsequent seeding) from a suspension to be a single passage, no matter how many aliquots you actually draw. Someguy1221 (talk) 20:32, 31 December 2007 (UTC)[reply]

What if I draw an aliquot every two days for 40 days. Have the 20 passages of my master culture been used? Probably not. But if I halve the culture instead of taking aliquots, then I have, right? --Seans Potato Business 20:41, 31 December 2007 (UTC)[reply]

I was just using aliquots since you mentioned it. It doesn't matter what container you're moving it in, or even that it's leaving a container. As far as my experience tells me, if you're suspending the cells at all, you've carried out a passage (I don't know what the precise rule would be for passaging cells that actually grow suspended). Unfortunately, I seem to have lost my guide on cell culturing (damnit). Someguy1221 (talk) 20:50, 31 December 2007 (UTC)[reply]

Decay Rates

This website [2] shows how decay times depend on the force responsible for the decay (scroll down). Is there any (hopefully simple) way to calculate this using the coupling constant, mass etc.? Thanks, *Max* (talk) 22:22, 31 December 2007 (UTC).[reply]