Wikipedia:Reference desk/Archives/Science/2010 March 12

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March 12

Looking for a bird

Does anyone know of a complete list of species featured in The Life of Birds? In particular, I am looking for one bird species, it must have been in "Signals and Songs" (episode 6). Its song sounded like early electronic analogue synthesizers, or someone rapidly turning a ham radio dial. I think Attenborough's commentary mentioned that it had to learn to produce something like 40 different sounds in order to get its song right. I think it was North American, small, and dull-coloured (mostly brown). Thanks in advance! ---Sluzzelin talk 01:22, 12 March 2010 (UTC)

Well no doubt the bird with the strangest vocals in that series was the Lyre bird, that's the one that can imitate other birds and even things like camera shutters and chainsaws, but I don't know if anyone's brought a Moog for it to imitate:) . However it's not American, and its tail is a big give away, but I distinctly remember that in a lot of close shots with the tail mostly if not totally cropped out it looks small and dull colored. Vespine (talk) 04:24, 12 March 2010 (UTC)

Thanks, I too remember the supreme superb lyre bird distinctly, but as amazing as it is, it isn't it. The one I'm looking for isn't mentioned in the article. It didn't imitate. It just had its entirely own spacey sound. :-) ---Sluzzelin talk 04:31, 12 March 2010 (UTC)

I think I found it. This page mentions the cowbird using "40 different notes, some so high we can't hear them". Judging by the sound sample on this site, it's the Brown-headed Cowbird. My memory had something slightly more spectacular in storage, but this is definitely it. Thanks again. ---Sluzzelin talk 06:05, 12 March 2010 (UTC)

(I realize I was the one asking, but maybe others are interested too): The Brown-headed Cowbird article doesn't mention its song, but I found something under Lateralization of bird song. ---Sluzzelin talk 12:24, 12 March 2010 (UTC)

Reflex subwoofer query

I made a ported (reflex) sub-woofer with a vertically upward firing drive unit and a vertically downward firing port for bass instrument reproduction. I found that increasing the distance between the port and the floor (to about twice the ports diameter) seems to give a much richer sounding bass then when I used only a single diameter spacing (as recommended in some sources as the min). Why is this, and how does the spacing between port and floor affect the response/efficiency of the system? In particular, why do larger spacings seem to give more 'fruity' bass with apparently increased high frequency response?--79.76.188.14 (talk) 01:47, 12 March 2010 (UTC)

Sound pressure waves emitted from the port are reflected by the floor so creating a resonant column of air. Raising the woofer increases the height of the column so lowering its resonant frequency. That reduces the "boxyness" of the sound in the same way as would enlarging the cabinet. The sound frequencies are too low for a floor carpet to provide effective damping but you might notice some improvement from filling the whole space between the port and floor with a roll of carpet felt or foam. Cuddlyable3 (talk) 11:08, 12 March 2010 (UTC)

I find myself wondering whether more of the sound emitted by a downward facing speaker doesn't end up in the apartment below you than in your own. StuRat (talk) 12:31, 12 March 2010 (UTC)

Depends how stiff the floor is.--79.76.188.14 (talk) 20:27, 12 March 2010 (UTC)

Likely outcome of an all-out thermonuclear war between the US and Russia at the height of their nuclear capabilities?

US and USSR nuclear stockpiles during Cold War

One of many possible fallout scenarios from a mid-1980s attack on the US

Obviously no-one can 'win' a global nuclear war, as such - but I remember reading somewhere that in the event of a full-on nuclear exchange where everyone threw everything at everyone else, it was hypothesized by that it would likely be that Soviet Union that would 'continue to exist' in some form in the aftermath (mainly due to its greater land mass and more dispersed population), whilst the US (and all of Europe - West and East) would be almost completely destroyed. Am I remembering correctly here? Does anyone know the piece of research I am referring to? Does this sound at all plausible? Thanks. --95.148.106.148 (talk) 02:22, 12 March 2010 (UTC)

Just for reference, according to that infallible source, Wikipedia, the peak of the cumulative US-USSR nuclear stockpiles was in 1986, at 63,977 total warheads (23,254 US, 40,723 USSR). That would also be a period of considerable technological sophistication in delivery vehicles—ICBMs, SLBMs, MIRVs. So that's a lot of destruction.

The main problem here has and continues to be that calculating that kind of destruction is pretty hard. Most estimates probably under-estimate the destructive power of the weapons—they exclude good predictions of fallout and fire effects, or the long-term effects of kicking up that much (radioactive) dust into the atmosphere (which would probably have effects on the climate as well). Lynn Eden's book Whole World On Fire discusses this at some length. Certain weapons effects are "easy" to model—blast pressure, prompt neutron radiation, gamma rays, etc.—and some are quite hard—firestorms, fallout, etc. As a result, most calculations focus on the "easy" bits and exclude the hard bits, even though the hard bits actually do a huge amount of the damage (most of the destruction in Hiroshima and Nagasaki was caused by fire, not radiation).

But I think it is clear that if the US were hit by a large nuclear exchange, depending on the weather patterns, the fallout itself would be pretty destructive to its ability to continue as any kind of a nation. It would also surely lose a huge, huge percentage of its population, clustered as they are in big cities on the coasts. The USSR would probably do a little better, though I'm not sure it would be all that enviable. The interior would surely be targeted (as would the US's) because that's where its factories and silos were (as with the US; note the FEMA fallout map picture that shows how hit the midwest is for this reason). The resulting fallout would coat the interior areas pretty good. The lands would not be productive without massive cleanup and long-term birth defects and etc. would be rampant amongst any survivors not living far underground in mythical Strangelove-esque fallout shelters.

If you pick an earlier point in time, when the stockpiles are not so huge and the delivery mechanisms are not so good, then the USSR does a lot better—bombers are not going to saturate the interior so much. The US also probably does a lot better, as Soviet delivery mechanisms were really quite poor until the late 1950s. --Mr.98 (talk) 02:41, 12 March 2010 (UTC)

I think another major issue would be the reliability of the payload delivery systems. Many articles I have read over the last few years suggest that many, many Russian delivery systems and payloads themselves are faulty, and would likely not launch correctly, if at all. Beach drifter (talk) 03:55, 12 March 2010 (UTC)

Is this referring to now, or then? I know that the Russian nuclear arsenal has supposedly fallen into disrepair somewhat since the end of the Cold War. Thanks for the answers so far, by the way. It boggles the mind that things ever got to the point where people needed to sit and calculate this stuff and that there were people who would be prepared to actually carry it out (or put the choice to do so in the hands of computers!) - it seems completely insane, not to mention quite obscene. --95.148.106.148 (talk) 04:03, 12 March 2010 (UTC)

Many people, including me, disagree on your last point, but this is not a discussion board. 74.212.140.226 (talk) 07:08, 12 March 2010 (UTC)

To clarify: The objective of having those weapons was as a deterrence. In order to deter, it was 100% essential that the enemy be utterly convinced that you'd push the button. Hence, even if you had no intention of ever doing such a crazy thing - you had to convince them that you would. Thus, at least the public face had to be that you'd do it. Although it pains me to say it - it worked. There was no world war three. The Soviets didn't invade Europe - despite having conventional forces that were easily capable of doing so. The US didn't extend it's power throughout the world in a way that the Soviets would have found unacceptable. The American people got so sick of the idea of war that it became difficult for them to carry through with them - the Russians ran out of money. We got through that period with nothing more than a few minor skirmishes - and it was actually a fairly peaceful time for the participants. The push that the cold war gave to technology has given us cheap computers, satellites, GPS and the Internet to name but a few. I don't like it - but it did work exactly as advertised and it had significant positive outcomes. The problems the world has now relate to the fact that one side of the conflict regards Mutually assured destruction as an acceptable outcome so deterrence doesn't work. SteveBaker (talk) 13:47, 12 March 2010 (UTC)

It's of note though that there is a fair consensus that the massive build-ups led to situations of overkill, well above and beyond what was probably necessary for deterrence (and there is a lot of evidence that the Cold War politicians and military people never actually really accepted deterrence as their goal—many were explicitly interested in the possibility of a debilitating first strike and feared that the other side was pursuing it as well). And there is a debate about whether creating the realistic possibility of mega-deaths really was preferable to the Soviets running Western Europe. I think there are many who would disagree that all of the skirmishes in that period were "minor"... a few million people died on all sides in Korea, Vietnam, Afghanistan, etc. At virtually no point were the Cold War powers not engaged in some kind of costly warfare. The idea that the Cold War powers acted as logical agents here is a somewhat naive reading of the history. It is luck as much as anything else that prevented nuclear exchanges, and there were some very, very close calls. --Mr.98 (talk) 16:40, 12 March 2010 (UTC)

I wouldn't say the Russian delivery systems are faulty. Russia has a better track record than the USA developing reliable rockets. Quest09 (talk) 10:21, 12 March 2010 (UTC)

It's really pretty hard to know if the delivery systems would have worked on either side. The US in particular only did component testing (they fired exactly ONE live rocket with a live warhead on it in the entire Cold War—and even that wasn't a stockpile warhead, if I recall). The essential problem is that you can't test whether everything would work correctly without actually using it, which means death. This problem and its historical consequences is discussed in depth in Donald MacKenzie's Inventing Accuracy: A Historical Sociology of Nuclear Missile Guidance, which is a really interesting book if you care about such things! --Mr.98 (talk) 16:40, 12 March 2010 (UTC)

One thing to consider is that the *vast* majority of US nuclear strategy was focused on destroying the Russian nuclear force, and presumably the Russians had a similar idea. This means that a total exchange would be very unlikely as each successful hit by one side means that many fewer warheads for the other side to launch. --Sean 13:28, 12 March 2010 (UTC)

How do you figure? Unless you do something to prevent the weapons launch, there will be a warning of 15 minutes or so for central missile silos. More then ample time to launch the weps, so if you target enemies silos, unless you have a way of masking your hundreds of launches, you are not going to destroy very many opposing warheads. Googlemeister (talk) 14:02, 12 March 2010 (UTC)

Nuclear weapons installations were the main targets of each side's nuclear forces. --Sean 16:30, 15 March 2010 (UTC)

Yes, I think your logic is wrong here. It's a "use 'em or lose 'em" scenario. There is enough time between launch and detection for the other side to launch theirs—that's why the big missile silos are out in the middle of the land masses (to give more time), and a considerable part of the force is kept in hard-to-detect submarines. That's why in the FEMA fallout diagram above, Montana, North Dakota, and Wyoming gets plastered so hard—they are assuming that in a full nuclear exchange, the Soviets would be seeking to really saturate the air force bases out there that were basing Peacekeeper and Minuteman missiles at the time. That does not necessarily mean at all that the Soviet hits would be successful in knocking out US capabilities—the missiles could easily pass each other on the way there. --Mr.98 (talk) 16:40, 12 March 2010 (UTC)

So you're suggesting the Soviets were planning on plastering those missile bases after the missiles were gone? That makes no sense. Counterforce -- knocking out the other guy's nukes -- is the reason very accurate missiles were considered destabilizing and were even banned in some treaty proposals and the reason SLBMs were considered (mostly) stabilizing (their missiles had lesser accuracy (implying countervalue strikes) and less ability to be countered since they were basically undetectable). --Sean 16:30, 15 March 2010 (UTC)

I'm suggesting that the Soviets would not have attempted to figure out which missile silos had launched and thus remove them from their target list. I'm also suggesting that they would have not had time to do this in a true second-strike scenario. --Mr.98 (talk) 18:45, 16 March 2010 (UTC)

A strange game. The only winning move is not to play. How about a nice game of chess? Coreycubed (talk) 15:20, 12 March 2010 (UTC)

I don't understand this discussion. An all-out nuclear war would, most likely, result in Nuclear winter, which would wipe out most of the world. Whether the US or USSR would fare better is completely irrelevant (after the first month or two). A 2007 study (using modern climate modelling techniques, which are pretty reliable) predicted that if all of the current stockpiles were launched, which are about 1/3 the size of the peaks, then the results would be:

A global average surface cooling of –7°C to –8°C persists for years, and after a decade the cooling is still –4°C (Fig. 2). Considering that the global average cooling at the depth of the last ice age 18,000 yr ago was about –5°C, this would be a climate change unprecedented in speed and amplitude in the history of the human race. The temperature changes are largest over land ... Cooling of more than –20°C occurs over large areas of North America and of more than –30°C over much of Eurasia, including all agricultural regions.

Those kind of drops would leave North America and Eurasia in Arctic conditions, there would be no significant agriculture. I haven't found estimates of death toll, but I would personally guess that we're talking at least 90% of the world's population being killed before the climate started to get back to normality (and that is assuming there isn't a Snowball Earth scenario where you get a runaway cooling which leads to thousands of years of global ice). --Tango (talk) 17:21, 12 March 2010 (UTC)

Assuming it is a USA vs. USSR war, the Southern Hemisphere would probably do much better than the North with respect to nuclear winter. Stratospheric wind patterns make it hard for dust injected in one hemisphere to cross the equator and reach the the other hemisphere (as shown by high latitude volcanic eruptions). Dragons flight (talk) 17:39, 12 March 2010 (UTC)

Contrary to the 1957 nuke fiction On the Beach (novel). In his book for some reason the fallout radiation levels do not decay to a very low level in a couple of weeks as most sources predict, but continue lethal for years. Edison (talk) 18:09, 12 March 2010 (UTC)

Indeed, radiation is a risk for those reasonably close to targets shortly after they are bombed. They may be a statistical increase in cancer risk for others, but that's about it. --Tango (talk) 19:38, 12 March 2010 (UTC)

If there is appreciable fallout in their food supply/land/water (as one would expect from a 1980s exchange of thermonuclears), the cancer and birth defect risks would be more than "statistical". --Mr.98 (talk) 01:56, 13 March 2010 (UTC)

That's true - as the quote says, it will be worst in North America and Eurasia. We can expect the southern hemisphere to fare better, but we're still talking a several degree temperature drop for years, which would be enough to ruin agriculture. Most of the population is in the north anyway. --Tango (talk) 19:38, 12 March 2010 (UTC)

I do not have the reference available, but I read a comment by someone who once was tasked with aiming US nukes at Russia, to the effect that it is easy to select the first several hundred targets, but by the time you are down to the 20,000th most important target, it is likely worth less than the bomb and its delivery system. 300 bombs could hit every US city with 100,000 people. 500 more could hit every military base in the U.S. Picture the Russian planner selecting target number 40,000: The courthouse in a small U.S. county? A 2 lane highway bridge? An electrical substation? A shopping center? Edison (talk) 18:03, 12 March 2010 (UTC)

A survey of basically all fiction about nuclear war, with some notes on its accuracy or lack thereof, by Paul Brians is available at [1]. Edison (talk) 18:21, 12 March 2010 (UTC)

Russia was known to have developed a crude anti nuclear missile technology (that the US may have duplicated too), which consisted of simply planning to detonate nuclear warheads in the direct path of incoming nuclear missiles. You don't have to be very precise that way. As I recall, the response to this technology was to plan to supersaturate certain targets, e.g. aiming 30 missiles at Moscow, so that at least one would get through and destroy the city. When you have a huge excess of warheads it is easier to make plans like that. Dragons flight (talk) 21:16, 12 March 2010 (UTC)

The Russian technology in question was the ABM-1 Galosh. It would have only been stationed around a few major cities. The American response was to ramp up MIRV development—make it so that there are 10 offensive missiles to every one defensive one. I don't think the US ever developed anti-missile technology of this sort, but they did have anti-bomber technology that worked this way (MIM-14 Nike-Hercules). Fun stuff. --Mr.98 (talk) 01:56, 13 March 2010 (UTC)

The non-negligible chances of missiles not launching cleanly and warheads not detonating also make it wise to send several to each target. Edison still has a point, though - even if you launch 30 at each target you still need about 1000 targets, which is more than enough to include every major city and military base. --Tango (talk) 21:32, 12 March 2010 (UTC)

Though it does get kind of ridiculous. In the 1960s (according to Richard Rhodes), Robert McNamara asked the air force how many missiles their force requirement calculations would say were necessary for Hiroshima. The response was that according to their up-to-date calculations, it would take three bombs of 80 kt each to shut down Hiroshima during World War II... or some 18X what was actually required to do "the job." The Eden book I cited earlier discusses in some detail the origins of these calculations and why they tended towards crazy levels of overkill. It is hard not to see some of this as just being about "we've got the money and can do it so let's do it" rather than serious questions about the necessity or consequences. Compare this with China's strategy, which is just about seeing how few missiles you can have and still maintain a viable second-strike deterrence—which they've concluded is about 200 or so. Much cheaper, less dangerous, and makes no pretensions of being a first strike force. (Which, for all that we talk about second-strike deterrence today, was in fact what the US policymakers were trying to do at the time, as the historical records have shown. Which is scary.) --Mr.98 (talk) 01:56, 13 March 2010 (UTC)

Speaking of games, this may be a subject relavent to game theory, zero-sum games and the Nash equilibrium. Shortly prior to the peak in Soviet warheads, the incident involving Korean Air Lines Flight 007 occured, and the Soviet Union may have planned a preemptive attack against the US in 1984. ~AH1^(TCU) 02:51, 13 March 2010 (UTC)

The US certainly planned for preemptive (first strike) attacks against the USSR all through the Cold War. There were a load of close calls: Cuban Missile Crisis, Stanislav Petrov, Able Archer 83, etc. The idea that things were perfectly rational, stable, and that military and political officials in the US and USSR actually took deterrence seriously is demonstrably historically false! It is the kind of pat reassurance given by TIME magazine, but it is not serious history. --Mr.98 (talk) 19:48, 13 March 2010 (UTC)

Just read the Stanislav Petrov article. So the scenario described in the song '99 Red/Luftballoons' nearly actually happened? Holy crap. --95.148.105.19 (talk) 22:34, 13 March 2010 (UTC)

For another list of "close calls", take a look at World War III. ~AH1^(TCU) 23:41, 13 March 2010 (UTC)

csi

whats a good article on here about evidence collection like dna fingerprints ect. in real life —Preceding unsigned comment added by 67.246.254.35 (talk) 05:42, 12 March 2010 (UTC)

Forensic science would be a good launching point; be aware that CSI is 99% bullshit. Its good theatre, but not good law work. Forensic science is a pretty good omnibus article on the topic. Much of CSI centers around the analysis of Trace evidence, but they also discuss many other aspects of forensics. Also see the CSI effect, a real problem in the forensic science world where people tend to believe that the stuff they see on CSI is somehow real, and it makes it hard for real forensic scientists to work with people like prosecutors and juries who have an unreal expecatation about what their job is like. (Full disclosure: My wife works in forensic science for a large police agency). --Jayron32 06:04, 12 March 2010 (UTC)

is it true they can take DNA from skin cells if someones arm or something brushes against an object? —Preceding unsigned comment added by 67.246.254.35 (talk) 06:33, 12 March 2010 (UTC)

Yes, but this also means that skin cells from many people are likely to be on any object. This makes it not so useful. You can say "Your skin cells are on the murder weapon", but they can say "So what, a dozen people's skin cells are on the murder weapon". StuRat (talk) 12:43, 12 March 2010 (UTC)

For lovers of CSI, I recommend: http://xkcd.com/683/ SteveBaker (talk) 13:31, 12 March 2010 (UTC)

This is also good. http://www.phdcomics.com/comics/archive.php?comicid=1156. --Mark PEA (talk) 16:21, 12 March 2010 (UTC)

orbit of moon

Is the expansion of the moon's orbit related to the expansion of the Universe? 71.100.11.118 (talk) 12:23, 12 March 2010 (UTC)

No, it's because the Moon is outside the Earth's geosynchronous orbit distance of 26,200 miles. Orbits below that distance will decay inward and those above that distance will decay outward, due to tidal interactions. StuRat (talk) 12:36, 12 March 2010 (UTC)

Does this explanation apply to the solar orbits of the planets? 71.100.11.118 (talk) 12:50, 12 March 2010 (UTC)

Yes, with all planets being outside that range, so having their orbits decay outward. However, the effect is tiny for the Moon, and even more so for the planets, as the tidal effects of the Moon on the Earth are far greater than the tidal effects of the planets on the Sun. Mercury, being the closest, would have the greatest effect, but even that might be too small to ever measure. StuRat (talk) 15:45, 12 March 2010 (UTC)

Also does this explanation mean that the moon has never orbited closer than 26,200 miles? 71.100.11.118 (talk) 12:54, 12 March 2010 (UTC)

Yes, unless something like a third body pushed it out of a closer orbit. StuRat (talk) 15:45, 12 March 2010 (UTC)

No, the rotation period of the Earth was originally ~8 hours / day, meaning geosynchronous orbit would have been much closer. Hence the moon also could have been closer (e.g. as close as 13000 miles). Most of the change in rotation rate is in fact attributed to the moon. Dragons flight (talk) 17:33, 12 March 2010 (UTC)

Good point. StuRat (talk) 19:10, 12 March 2010 (UTC)

The second paragraph at Orbit_of_the_Moon#Tidal_evolution_of_the_lunar_orbit is a very nice explanation of how the tidal bulges on the Earth coupled with the Earth's rotation cause the moon's orbit to increase. Does that help? Zain Ebrahim (talk) 13:03, 12 March 2010 (UTC)

So the interaction of Earth's spin, tides, are entirely responsible for expansion of the moon's orbit rather than the expansion of the Universe. 71.100.11.118 (talk) 13:20, 12 March 2010 (UTC)

Yes, Expansion of the universe has zero effect on the orbit of the moon. Dauto (talk) 13:28, 12 March 2010 (UTC)

Interestingly, by calculating how much the Moon's distance from Earth should increase if Hubble's law would be correct for such small distances, I get 28 mm/year, on the same order of magnitude as the 38 mm/year in measured increase of the distance, but I guess that is coincidence. Icek (talk) 18:05, 12 March 2010 (UTC)

I just did that calculation too (having not refreshed the page since you posted) and concur. It is a pretty amazing coincidence, but it is a coincidence. Our understanding of tides (which are based on very simple physics) predicts a recession of very close to the measured amount, so that understanding would have to be totally wrong for it to be caused by anything else. If our understanding of tides is wrong then our understanding of cosmology doesn't stand a chance of being right. --Tango (talk) 18:30, 12 March 2010 (UTC)

is there intelligent sleep like beauty sleep?

If you google beauty sleep, you get a whole lot of hits, at least some of them quite credible. Moreover, if you've ever seen the harried look of someone who's gone with very little sleep over many days, it is obvious that they're not quite well. It's ugly. Now, I wonder if there is an analogous effect on intelligence. Is someone intellectually "harried" by prolonged insufficient sleep? Do they absorb things they learn less, and so forth? I have a more specific question: when we say "sleep deprivation" we mean considerably less than 6 hours of sleep per day, maybe as little as 2-3 hours per night. (By the way, if anyone wants to edit the sleep deprivation article, it certainly doesn't define the amount of time of chronic sleep deprivation. Is it less than 3 hours per night, less than 4, than, 5, than, 6, than, 7, or what?)

Anyway I have a specific question. When people talk of BEAUTY SLEEP they don't mean "don't be sleep deprived", ie don't sleep just 3 hours. What they really mean is: don't sleep too little, only 5-6 hours. Sleep 8 or 9 hours instead.

Now, insofar as this works at giving people a healthier look, I wonder: does it have an intellectual effect as well? Does it make sense likewise to talk of 'intelligence' sleep like 'beauty sleep'.

Specifically, I know that it is hard to absorb and digest information if you are not able to sleep on it, severe, chronic sleep deprivation of only 2-3 hours of sleep per night wreaks havoc on one's cognitive capabilities...

but is there an EXTRA intellectual benefit, like the beauty benefit, to sleeping, say, 9 hours per night while learning a lot, as compared with only sleeping 6? Where is the cutoff on the return? Surely 12 hours per night doesn't make you any better at absorbing information and so forth than 9 hours per night, but surely 6 hours is much better for you with respect to the same cognitive effects versus 2 hours... so, is there a graph someone could show, or the specific cutoff points on the return?

I can imagine several possible graphs: I wonder which one is correct. Thank you. —Preceding unsigned comment added by 82.113.106.93 (talk) 13:09, 12 March 2010 (UTC)

I hope someone turns up with some studies to cite soon, but until then I can tell you from experience that there is such a thing as being intellectually not-at-full-strength because you don't get quite enough sleep for days on end: talk to some parents. I can also tell you that you're unlikely to find graphs in the form you want that can be generalised, because different individuals have different sleep needs, and this can vary depending on circumstance. For example: when I start a new job, I find I need more sleep for the first week or two. 86.178.167.166 (talk) 01:25, 13 March 2010 (UTC)

Not sure about specific studies, but some relavent articles include Sleep#timing, sleep and creativity, lucid dreaming, delayed sleep phase disorder, insomnia, psychomotor learning and health and intelligence. ~AH1^(TCU) 02:33, 13 March 2010 (UTC)

According to a book I read by a sleep doctor (can't recall the title), 95% of people require 8 hours of sleep per day on average- and that less sleep will lower your efficiency. It complicates things, because a person low on sleep sleeps DEEPER, but stills needs to make it up later. (By the way, the other 5% requires as little as 6 hours total to as much as 10 hours.) If I remember the title, I'll post it. Mxvxnyxvxn (talk) 23:16, 16 March 2010 (UTC)

Lorentz invariant

If two inertial systems (x, y, z, t) and (x', y', z', t') are moving relative to each other, with their origins coinciding at t = t' = 0, and at that moment a light pulse is emitted from the origin, then x² + y² + z² - c²t² = x'² + y'² + z'² - c²t'² = 0. To show that x² + y² + z² - c²t² = x'² + y'² + z'² - c²t'² is true in general, my textbook just applied the Lorentz transformations to (x, y, z, t) and crunched the algebra. Is there a simpler way (that is, by considering some sort of thought experiment or something) to show that x² + y² + z² - c²t² is invariant under a change of reference frame? —Preceding unsigned comment added by 173.179.59.66 (talk) 14:40, 12 March 2010 (UTC)

The book did that calculation just to double check the consistency of the theory. In fact that equation is nothing more than the statement that the speed of light is the same for all observers. This is the starting point of the theory and is usually taken as a postulate. No gedanken experiment is necessary. Dauto (talk) 17:04, 12 March 2010 (UTC)

After edit conflict:

Sure. Imagine the event E with coordinates (x,y,z,t) and (x',y',z',t'). Let's call the common origin event O. Let's work with c=1 and let's forget about y and z. I assume that's what the textbook does as well. (which book do you have?)

First assume t^2 > x^2. Assume a clock present at events O and E, which was also set to t"=0 at O. The unprimed/primed systems have a speed v resp. v' w.r.t. this clock. Then clearly v^2=x^2/t^2 and v'^2=x'^2/t'^2. Through the time dilation equations, the proper time t" on that clock satisfies t = t" / sqrt(1-v^2) and t' = t" / sqrt(1-v'^2). This implies that t" = t sqrt(1-v^2) = t' sqrt(1-v'^2). Substituting the values for v^2 and v'^2, we get the result.

Then assume t^2 < x^2. Assume a rod stretched between events O and E, and a system in which this rod is at rest. The unprimed/primed systems have a speed v resp. v' w.r.t. this system. (these v and v' differ form the previous ones!) Then clearly (see spacetime diagram!) 1/v^2=1-x^2/t^2 and 1/v'^2=1-x'^2/t'^2. Through the length contraction equations, the proper lenght x" of that rod satisfies x" = x / sqrt(1-v^2) and x" = x' / sqrt(1-v'^2). This implies that x / sqrt(1-v^2) = x' / sqrt(1-v'^2). Substituting the values for v^2 and v'^2, we get the result again. DVdm (talk) 17:45, 12 March 2010 (UTC)

Thanks, and I have Intro to Mechanics (Kleppner/Kolenkow). Oh and to Dauto: I don't think the Lorentz invariance thing is a statement that the speed of light is constant with respect to different inertial reference frames. The (x, y, z) coordinates don't have to be the point occupied by a light particle, just any point in general, so I don't think that you can extrapolate that because x² + y² + z² - c²t² = x'² + y'² + z'² - c²t'² = 0 for a beam of light, x² + y² + z² - c²t² = x'² + y'² + z'² - c²t'² is true for any event (x, y, z, t). If I'm wrong, please let me know! 173.179.59.66 (talk) 17:57, 12 March 2010 (UTC)

That's the reason why I elaborated. By the way, this Kleppner/Kolenkow is a very good textbook. Stick with it :-) - DVdm (talk) 18:43, 12 March 2010 (UTC)

You are correct, but what I said is also correct. There is no need for a gedanken experiment showing that the relativistic interval $dS^{2}=(cdt)^{2}-dx^{2}-dy^{2}-dz^{2}$ is an invariant. This is a postulate of the theory. In other words, you can define the Lorentz transformations as the set of transformations that respect the invariance of the relativistic interval the same way that you can define rotations as the set of transformations that respect the invariance of space-vector lengths. Dauto (talk) 18:42, 12 March 2010 (UTC)

Oh I see, thanks. 173.179.59.66 (talk) 02:23, 13 March 2010 (UTC)

I agree, but in many (older) books the invariance is not postulated, but rather derived from the "classic" postulates. DVdm (talk) 18:49, 12 March 2010 (UTC)

do you retain information better if you sleep on it?

this is related to the question above. My question is: do you retain information better if you've slept on it versus being tested the same day?

I don't mean because of the additional time since you learned it: say you compare learninng at 8 AM and being tested same day at 8 PM versus learning at 8 PM and being tested the next morning at 8 AM. Thank you. 82.113.121.104 (talk) 17:46, 12 March 2010 (UTC)

Several recent studies indicate this is the case. Even an afternoon nap can help. Graeme Bartlett (talk) 17:54, 12 March 2010 (UTC)

How did they control for tiredness? --Trovatore (talk) 17:55, 12 March 2010 (UTC)

According to Tony Buzan, your recall improves shortly after you have studied something (because your brain has sorted the information out) and then rapidly falls off. He recommender (and I found this works well) is to review the learn material 10 minutes after studying it, then review it again 24 hours later. The again after a week and once more after a month. This helps the information to be retained by the ‘long term’ memory. So, back to your question. You are likely to get higher marks if tested the same day. But it is better to keep reviewing it if you what to remember it long term. From Page 54, 55 ,56. Use your Head by Tony Buzan.--Aspro (talk) 18:21, 12 March 2010 (UTC)

I'm wondering too. I would like to test if this is the case for me. Guys here is my followup question:

can someone suggest a test methodology to control for everything but having "slept on it"

I would like to test if the above is true for me, and although the sample size will be 1, nevertheless that 1 subject is the one of greatest relevance to me :).

So can someone suggest an appropriate test methodology for me, for the simple task of trying to memorize and retain 10 of the multiplications from the 1000 x 10000 multiplication table, for example,

328x340=111520
274x781=213994
163x310=50530
494x65=32110
491x753=369723
969x954=924426
667x324=216108
77x813=62601
622x647=402434
196x249=48804

How should I do this in a way that will control TIREDNESS, control the stress or interference of other aspects of my life (e.g. I might concentrate better at night, since I wouldn't be doing something better anyway, but in the morning, I might be stressing out over something more important that I should be doing instead, and not concentrate fully and so forth). Any ideas? Thanks. 82.113.121.104 (talk) 18:09, 12 March 2010 (UTC)

You can't possibly do a well-controlled experiment with yourself as subject. The most common experimental paradigm is to have two groups of subjects, let's call them A and B. Group A learn the task, take a nap immediately afterward, sleep as much as they want to during the night, and are tested on the following day. Group B do the same things except without the nap. Thus both groups have the same learning experience and are well-rested during testing -- the only difference is the nap. For a recent review of the literature pertaining to this topic, you might look at PMID 19251443, if you have access to it. Looie496 (talk) 19:03, 12 March 2010 (UTC)

Obviously I can't do a double-blind experiment with myself as a subject; that's a given. But why couldn't I do a well-controlled experiment with myself as a subject, granted it might take longer, but you can see from my methodology below that you can still have a control...

Now my deeper problem with the methodology you mention is that BOTH groups a and b get to "sleep on it"!!! I am interested in the difference between being able to sleep on it, versus not having slept between learning the material and being teested.

To this end, what about the following test methodology:

For the next 20 days, every morning I get up at 6, every morning at 8 I put aside 15 minutes for the Methodology, every night I put aside 15 minutes at 8 PM for the Methodology and go to sleep at 10 PM.

• I group my these 20 days into groups of two (call them day1 and day2 within each group). At the onset, I randomize each pair of days into either being a "group a" pair of days or a "group b" pair of days.
• Then, each day goes as follows:
• If it is day1 of a group a, I do the studying in the morning slot, and take my test in the evening slot.
• If it is day2 of a group a, then I don't do anything in either the morning or evening slot.
• If it is day1 of a group b, I don't do anything in the morning slot. In the evening slot I do the studying.
• If it is day2 of a group b, I take the test in the morning slot. I don't do anything in the evening slot.

Now what could possibly be wrong with that methodology? It has both a control (group a pairs of days and group b pairs of days) and beyond this, it has an equivalent 12 hours between the studying and the being tested - the only difference is one of the groups includes 8 hours of sleep in that 12 hours, and the other group does not. Now, if there is a statistically significant difference in this well-controlled study on myself, it can be argued that the difference is because I am "more tired" at 8 PM than 8 AM, or alternatively phrased, "more alert" at 8 in the monring, after having awoken at 6 AM. Also the effects of any coffee I would drink would have to be normalized (ie I would not drink coffee until after hte test in the mornning). Same for taking my showers. Is there any other way in which the methodology I've just proposed would fall short? Thank you. 82.113.121.104 (talk) 19:46, 12 March 2010 (UTC)

The problem is that test performance is influenced by motivational factors, and it's impossible for you to ensure that you're equally motivated to perform well under all conditions. Even if you don't realize it, you're sure to have some motivation to want one outcome rather than another. Looie496 (talk) 20:07, 12 March 2010 (UTC)

so what do I have to do? Do the test with my brother, but not tell him what I'm testing, ie he doesn't know if it's significant whether the recall test is the next day or the same day. Instead, he just thinks it's a multiday test? I guess here you're going to tell me that that's good, but not good enough: it is single-blind. Instead, I need to get my brother to do the test with someone else, WITHOUT letting my brother know that what is really being tested is whether the subject has slept on it. Obviously all this machination just to get the result "no difference" is ridiculous, when I might get the same result myself anyway, so you guys must really be convinced there's a big difference... :) 82.113.121.104 (talk) 22:51, 12 March 2010 (UTC)

If you want a real study, you need more than one subject. You need, at a minimum, a control group and a test group. By just testing your brother, you are getting nothing more than an anecdote - which is not data. Further, your statement appears to suggest that you think everyone here is involved in a conspiracy to keep a secret from you. That is silly. -- kainaw™ 22:56, 12 March 2010 (UTC)

Well, you'd need multiple subjects if you were doing this for publication, or even for a school project. If you're just trying to get an answer for yourself, you could do it by repeatedly testing a single subject -- but even then it's important that the subject not be aware of what you are looking for, in order to avoid having your subject subconsciously sway toward the result you want (or against it!). Double-blind experiments are always nice, but generally not possible in sleep studies -- you can't blind somebody to the difference between being awake and being asleep. Looie496 (talk) 23:21, 12 March 2010 (UTC)

Referring to your test numbers, some numbers have special properties including 142857. Another subject related to functioning versus sleep is sleep and creativity. ~AH1^(TCU) 02:11, 13 March 2010 (UTC)

There is also the question of what you mean by "retain" information, obviously you could take this as being able to repeat a set of numbers. But for something like understanding how a set of variables interact, sleeping can help you process the data, hence why you sleep on a decision. Now in this case you might "retain" the meaning of the information better but not be able to write down the table, equation etc precisely. 82.132.139.214 (talk) 02:20, 13 March 2010 (UTC)

center of mass

Technically, isn't it the Earth/Moon center of mass that orbits the sun making the Earth closer or further away from the Sun at times during the rotation about the center of mass? 71.100.11.118 (talk) 18:24, 12 March 2010 (UTC)

Yes, but the Earth is so much larger than the Moon that the centre of mass of the two is inside the Earth. That means the Earth wobbles a bit, but nothing more. --Tango (talk) 18:40, 12 March 2010 (UTC)

And this wobble (a few thousand km ?) is insignificant compared to the distance caused by the Earth's slightly elliptical orbit, which is on the order of 5 million km. StuRat (talk) 18:58, 12 March 2010 (UTC)

Deriving the Lorentz transformations

It seems that the common approach to obtaining the equations for the Lorentz transformations are by guessing at their form and then, by considering four seperate situations, determining the constants. From these equations, things like time dilation and length contraction can be worked out. Now, my goal was to go the other way around: starting from time dilation and length contraction, arrive at the Lorentz equations.

Suppose that in the reference frame O, the reference frame O' is moving at a speed v in the x-direction, with their origins coinciding at t = 0. An event E occurs at (x, y, z, t) in the O frame. It's straightforward to show that y' = y and z' = z. Next I considered x'. In the O frame, the distance between O' and E is x - vt. Because the ruler O' uses is shortened by a factor of γ , she will then measure the distance x - vt as being greater and O measures it, by a factor of γ. Thus, x' = γ(x - vt).

However, I'm having trouble with t'. I know that t' = γ(t - vx/c₂). I assume that the -vx/c₂ term comes from that, because O' believes that she's at rest, when the light emitted from the event reaches her, she doesn't treat herself as moving into the light, and thus there's a discrepancy as to how long before the light reaches O' did the event actually happen. Unfortunately, I can't arrive algebraically at this term. Finally, the gamma factor. I assume this comes from time dilation, but the wouldn't O' 's clock be running slower? So wouldn't the term have to be 1/γ? Can someone please tell me how to get the final Lorentz term this way? I know there are probably other, easier routes, but for personal reasons I would like to know how to do it this way. Thanks a lot! 173.179.59.66 (talk) 18:31, 12 March 2010 (UTC)

It seems that you're failing to take into account that the two observers perceive different simultaneities.Dauto (talk) 18:50, 12 March 2010 (UTC)

You might be interested in Bondi k-calculus. As I recall your desired approach is pretty much exactly that. Although the article isn't exactly detailed. 129.234.53.144 (talk) 19:43, 12 March 2010 (UTC)

Hmmm not really, Bondi starts from the Doppler shift and works from there, instead of from time dilation (his equations have a bunch of k's). And as for the simultaneity comment, I thought that by considering how long it would take the light to reach the observer in each frame, simultaneity would be accounted for (as that is how differences in simultaneity arise). And then there's the issue with the gamma factor, which I think should be 1/γ. Sorry, I'm still lost! 173.179.59.66 (talk) 04:01, 13 March 2010 (UTC)

"...the distance between O' and E is x - vt." => That is a distance between a pair of simultaneous events in the unprimed system. So these events are not simultaneous in the primed system. Then you arrive at "Thus, x' = ?(x - vt)." But that is a coordinate of the event E in the primed system. As such it denotes the distance between a different pair of events. When you draw a spacetime diagram, you see what is going on. A piece of advice: always think in terms of events, and always draw a spacetime diagram. DVdm (talk) 11:57, 13 March 2010 (UTC)

So then, how would you fix this conundrum with t'? 173.179.59.66 (talk) 12:16, 13 March 2010 (UTC)

I don't think there's anything to fix, and we surely haven't fixed the trouble with x'. We merely established that not only you are "having trouble" with t', but with x' as well. DVdm (talk) 13:12, 13 March 2010 (UTC)

The point (0, 0, 0, t - vx/c²) is the point along the O frame path that's simultaneous to E as viewed from the O' frame. In the O' frame, clocks in the O frame run slower by a factor of γ. To the O' observer, when event E happens, the clock in the O frame reads t - vx/c², and so the amount of time that has elapsed in the O' frame is actually γ(t - vx/c²). Rckrone (talk) 20:30, 13 March 2010 (UTC)

So basically, what you're saying is that we must look at the perspective of O', not O. Why would this be (afterall, we're starting with O 's coordinates, and the time dilation/length contraction all occur in O' 's frame, but this approach seems to yield the wrong answer). Sorry if I'm being a bit slow, I really want to understand this, subtleies included. 173.179.59.66 (talk) 21:16, 13 March 2010 (UTC)

We're trying to find the O' time coordinate of E, which is the amount of elapsed time that has been experienced by O' at the point where O' views event E as being simultaneous to her. So we need to consider the points simultaneous to E in the O' frame and how much time has passed in the O' frame up to those points. We already know what E looks like in the O frame: it's at point (x, y, z, t). Rckrone (talk) 21:32, 13 March 2010 (UTC)

Exactly. That's why I mentioned earlier that 173.179.59.66 was failing to take the different simultaneities into account. There is a way to avoid having to deal with simutaneities and time dilations. I'll post it later because I'm busy right now. Dauto (talk) 22:43, 13 March 2010 (UTC)

Things are making much more sense now, I eagerly await your posts! PS: how did you get the vx/c2 term?173.179.59.66 (talk) 01:18, 14 March 2010 (UTC)

Okay, I've made some headway myself, and have managed to understand why it should be t' = γ(t - ...) instead of t' = (1/γ)(t - ...). However, the vx/c² term still eludes me. Here's the work I've done: I looked at the O' frame, with the goal of getting (x,y,z,t) in terms of (x',y',z',t'), which can then be easily switched to get (x',y',z',t') in terms of (x,y,z,t). I should get t = γ(t' + vx'/c²). Now, the gamma factor comes from clocks running slower in the O frame relative to the O' frame. The vx'/c² should come from difference in percieved signal delay between reference frames. In the O' frame, the light from event E reaches O at t' + x'/(c-v). From O 's perspective, the signal delay should only be x/c, so she will conclude that E happened at t' + x'/(c-v) - x/c. x' = x/γ, so this becomes t' + x'(1/(c-v) - √1-v²/c²/c). Problem: this doesn't seem to algebraically reduce to t' + vx/c². Where's my mistake? 173.179.59.66 (talk) 03:03, 14 March 2010 (UTC)

To arrive at the expression for $x'\,$ we have to take two things into account - A space dilation/contraction and an offset. We can for instance assume that there is a ruler of length $l'=x'\,$ at rest with respect to the primed coordinate system $S'\,$ so that one end of the ruler coincides with the coordinate origin at all times and the other end coincides with the event that happens at $(t',x')\,$. Now, that ruler is at rest at the $S'\,$ coordinate system so $l'\,$ is its proper length. The length of the ruler on the unprimed coordinate system $S$ gets space contracted, $l={\frac {1}{\gamma }}l'$. That was the space dilation/contraction. The offset comes about because ot the relative motion between the two coordinate systems. From the point of view of $S\,$, by the time $t\,$ at which the event $(t,x)\,$ happens the origins of the two systems will be already a distance $d=vt\,$ apart and that distance - the offset - must be added to the length of the ruler to get the coordinate $x\,$, giving us the equation $x=d+l=vt+{\frac {1}{\gamma }}l'=vt+{\frac {1}{\gamma }}x'$. That last equation can be solved for $x'\,$ and we find $x'=\gamma (x-vt)$.

It so turns out that in order to arrive at the expression for $t'\,$ we also must take into account a time dilation/contraction and an offset. We can place a watch at rest with respect to system $S\,$ at the coordinate $x\,$ and start the watch exactly at the instant the two origins coincide (from the point of view of $S\,$). The watch will mark the time $T=t\,$ when the event $(t,x)\,$ happens. Now, that watch is at rest at the $S\,$ coordinate system so $T\,$ is a proper time, and this time interval will be seen time dilated from the point of view of the system $S'\,$. That is $T'=\gamma T$. That was the time dilation/contraction. The offset comes about because from the point of view of the system $S'\,$ the watch was not started simultaneously with the time the two origins coincided. There is a delay $D'\,$ given by $D'=\gamma {\frac {vx}{c^{2}}}$. $D'\,$ must be subtracted from $T'\,$ to get the coordinate $t'\,$. We get then the equation $t'=T'-D'=\gamma T-\gamma {\frac {vx}{c^{2}}}=\gamma (t-{\frac {v}{c^{2}}}x)$

Even though the space offset due to relative motion is very intuitive, the time delay due to non-simultaneity isn't. It is to our advatage if we manage to find the transformation without ever having to mention simultaneity or time dilation. To do that we start from the last equation of the first paragraph found using space contraction and relative motion only

$x'=\gamma (x-vt)$

We also have a similar equation obtained for $x\,$ in terms of $x'\,$ and $t'\,$. It's the same equation but with a sign difference because from the point of view of $S'\,$, $S\,$ is moving in the oposite direction

$x=\gamma (x'+vt')$

Substituting $x'\,$ from the first equation into the second,

$x=\gamma [\gamma (x-vt)+vt']$

$x=\gamma ^{2}(x-vt)+\gamma vt'$

$\gamma ^{-2}x=(x-vt)+\gamma ^{-1}vt'$

$vt+(\gamma ^{-2}-1)x=\gamma ^{-1}vt'$

$vt+[(1-{\frac {v^{2}}{c^{2}}})-1]x=\gamma ^{-1}vt'$

$\gamma ^{-1}vt'=vt+(-{\frac {v^{2}}{c^{2}}})x$

$\gamma ^{-1}t'=t-{\frac {v}{c^{2}}}x$

$t'=\gamma (t-{\frac {v}{c^{2}}}x)$

Voilla. Dauto (talk) 03:38, 14 March 2010 (UTC)

Wow that was exactly what I was looking for, thanks a lot! Really, that was a great help. I just have one final question though (sorry!): what was wrong with the signal delay approach I used above? I'm guessing it's another issue with simultaneity, but I just can't see were the problem is (or how it can be resolved, if I was inclined to use the signal delay way to avoid having to imagine clocks placed along the x-axis). I know it may seem trivial, but I want to have a thorough understanding of special relativity, and if I can't see why a certain approach is flawed then I can't say I really do. 173.179.59.66 (talk) 04:18, 14 March 2010 (UTC)

Never mind, I figured it out: I was multiplying my gammas at the wrong time. Much thanks again! 173.179.59.66 (talk) 05:35, 14 March 2010 (UTC)

Dauto, at a certain point you say: "There is a delay $D'\,$ given by $D'=\gamma {\frac {vx}{c^{2}}}$", I don't see how you arrive at this expression. Can you explain?

Perhaps I should add that I object to the explanation you have given above, as you have simply taken an equation from the inverse Lorentz transformation ($x=\gamma (x'+vt')$). Of course we all know that given any two of the four (linear) equations, the other two equations be derived algebraically. But 173.179.59.66 asked for an alternative way to derive the transformation: "...starting from time dilation and length contraction, arrive at the Lorentz equations.". I don't think that's what you did here. - DVdm (talk) 10:51, 14 March 2010 (UTC)

I didn't arrrive at the expression for $D'\,$. I took it as a postulate. The point of that paragraph was to highlight the similarities of the expressions for $x'\,$ and $t'\,$. I agrre that taking the expreesion for $D'\,$ as a postulate is not entirely satifactory and that's why I showed an alternate way to solve the problem on the third paragraph. The expression for the inverse transformation (($x=\gamma (x'+vt')$) can be obtained from space contraction in a manner similar to the one I used in the first paragraph to obtain the expression for $x'\,$, or better yet, simply use the principle of relativity and obtain the expression for $x\,$ from the expression for $x'\,$ by making the replacements $x'\rightarrow x\,$, $x\rightarrow x'\,$, $t\rightarrow t'\,$, and $v\rightarrow v'=-v$, afterall from the point of view of $S'\,$ it is $S\,$ that is moving. Dauto (talk) 15:52, 14 March 2010 (UTC)

Entirely with you now. Cheers - DVdm (talk) 16:01, 14 March 2010 (UTC)

tidal forces

For two absolutely solid bodies (such as pure diamond crystal matrix) orbiting each other and assuming no tidal forces then the spin of either body results in no geosynchronous orbit such that the orbit neither decay inward or outward regardless of the orbital distance, is that correct? 71.100.11.118 (talk) 18:33, 12 March 2010 (UTC)

You can still get friction and gravitational waves, which can remove angular momentum away from your system. Graeme Bartlett (talk) 18:48, 12 March 2010 (UTC)

The assumption that diamonds don't undergo tides is incorrect. Any matter will be affected by gravitational forces, and thus undergo tides, although solids deform much less than fluids. StuRat (talk) 18:53, 12 March 2010 (UTC)

There will always be tidal forces. I suppose if the body is sufficiently rigid that those forces don't result in significant bulges then we can discount them, though. In that case, there won't be any tidal interactions resulting in decay, in any direction for any orbit. There are other factors that will change the radius of the orbit, though - atmospheric drag, for instance (particularly for low orbits), and interactions with other objects. --Tango (talk) 18:57, 12 March 2010 (UTC)

Animals that feed only on animals of their own kind

Which animals eat nothing but other animals of its own kind? I know there are lots of animals that practice cannibalism but all cannibal species I think of usually eat other stuff too and I can't even think of any cannibal species that feeds primarily (let alone exclusively) on the animals of its own kind. I am aware that such behaviour would be very inconvenient for the survival of the species but some animals routinely eat their young and still thrive. Surtsicna (talk) 19:47, 12 March 2010 (UTC)

It's pretty obviously impossible -- since digestion is less than 100% efficient, the biomass of such a species would have to steadily decrease. Looie496 (talk) 19:59, 12 March 2010 (UTC)

Exactly. There would be no way for energy to enter the species. Even with 100% efficient digestion, not all that energy is converted to biomass. Much of it is used for locomotion, for example. --Tango (talk) 20:05, 12 March 2010 (UTC)

Agreed. You could, however, have a small portion of a species which cannibalizes the rest. I would expect that this portion would then evolve into a new species. StuRat (talk) 20:07, 12 March 2010 (UTC)

Whether they evolve into a new species would depend on whether they breed with the rest of the population or not. If they do, there won't be any speciation. It is common for, for example, dolphin pods to have different diets even in the same environment, but they still breed more generally. --Tango (talk) 20:42, 12 March 2010 (UTC)

I wouldn't expect much interbreeding between hunters and prey, both because the prey will run and hide and because the hunters will kill the prey if they find them. StuRat (talk) 16:00, 13 March 2010 (UTC)

Hmm. That argument implies all biomass comes from ingestion. What about if the animal were symbotic with something photosynthesising which lived in them? Not that I can think of any such but I am not clear it is technically impossible for an animal with a second such source of biomass to ingest only others of the species. --BozMo talk 21:14, 12 March 2010 (UTC)

In that case, it would have to either derive most of its energy from the other source or reproduce faster than it can eat. The first satisfies OP's question in letter but not in spirit - it would not technically be "eating" if it were to derive energy through photosynthesis. The second is unlikely given the normal ratio of consumption of nutrients to species reproduction found in other animals. That is to say, in my experience most creatures eat more often than they reproduce. Coreycubed (talk) 21:31, 12 March 2010 (UTC)

Well biology is not my subject but there are animals as far as I can tell which eat nothing (eg a Giant tube worm which derives its energy from non digestive processes and has no digestive tract). So variants could conceivably exist which ate only its own kind. But you have to be a long way out of the ordinary, and certainly not close to your ordinary ratios. --BozMo talk 21:51, 12 March 2010 (UTC)

Yes, but as Coreycubed says, that only satisfies the letter of the OP's question, not the spirit. --Tango (talk) 00:05, 13 March 2010 (UTC)

I seem to recall a certain species of spider where the young eat their mother for nutrition, probably Segestria florentina. Spider cannibalism is also relavent but those species do not solely depend on cannibalism for foon. ~AH1^(TCU) 02:06, 13 March 2010 (UTC)

Interesting, but I'm sure they eat more than just their mother during their lives. --Tango (talk) 02:16, 13 March 2010 (UTC)

Ants have three or more genders. If the workers, or whichever ones are non-reproductive, went around fattening themselves up and then the soldiers and queens ate them it could work like that, in the sense of enough energy gathered. ~ R.T.G 02:56, 13 March 2010 (UTC)

I wouldn't say they had more than two genders - workers are female, just infertile females. Also, I think soldiers are a type of worker. The males (drones) tend to do very little. All of that aside - what you describe would work, but it doesn't meet the OP's requirements since the workers are eating something other than their own species. --Tango (talk) 03:10, 13 March 2010 (UTC)

Terminal cancer

If someone is in the final stages of terminal cancer, why is nutrition denied?--79.76.188.14 (talk) 21:51, 12 March 2010 (UTC)

You need to be careful with terminology here, as nutrition and food aren't the same things. Nutrition is essential to survive, so to keep the patient comfortable they would need to be feel at least partially nourished, otherwise the individual will starve. As far as I know, in terminal cancer (i.e. where there will be no further medical intervention other to make the patient comfortable), food wouldn't be denied to the patient, but rather the patient may be unable to take food orally because of severe weakness associated with late stage cancer. Someone feel free to correct me there. Regards, --—Cyclonenim | Chat  23:04, 12 March 2010 (UTC)

Sorry I may have misunderstood your question, are you referring to nutrition being denied by the medical professionals, or nutrition to cells being reduced and why that occurs? Regards, --—Cyclonenim | Chat  23:22, 12 March 2010 (UTC)

Nutrition denied by medical professionals--79.76.188.14 (talk) 00:03, 13 March 2010 (UTC)

In that case, as Cyclonenim says, it simply isn't. Where have you got that from? Palliative care certainly involves feeding patients if they are willing and able to eat (and giving them intravenous nutrition if they can't/won't eat and it is appropriate to do so). --Tango (talk) 00:09, 13 March 2010 (UTC)

No, I think it is sometimes. It's a form of passive euthanasia. Generally we're talking about a patient in a coma or near-coma state, and who, if he did regain consciousness, would not likely enjoy it. --Trovatore (talk) 00:12, 13 March 2010 (UTC)

True, it is used for coma patients (although not usually in cases where they think there is a significant chance of the patient waking up). Do cancer patients typically end up in comas, though? Brain tumours sometimes cause comas, but I wouldn't expect other forms of cancer to. --Tango (talk) 00:55, 13 March 2010 (UTC)

When the body is unable to absorb nutrients due to organ failure but the nutrients continue to be supplied to the patient, conditions such as edema can occur. See also medical ethics. ~AH1^(TCU) 02:01, 13 March 2010 (UTC)

True, but by that point the patient would be at most hours from death anyway, wouldn't they? Not feeding someone for a few hours isn't unusual, even if they are perfectly healthy. --Tango (talk) 02:13, 13 March 2010 (UTC)

Some cancer patients lose the ability to process oral food weeks or even months before they ultimately die. Fluids and some nutrition can be supplied intravenously for a time, but the lack of solid food and difficulty absorbing nutrition can contribute to the sense of wasting away that is not uncommon in end stage cancer patients. Dragons flight (talk) 19:56, 13 March 2010 (UTC)

Absolutely, but that isn't nutrition being withheld. --Tango (talk) 21:22, 13 March 2010 (UTC)

Nutrition is sometimes withdrawn from terminal patients, with the consent of the relatives, in certain jurisdictions, as a legal method of euthanasia. The logic is that the patient "dies on their own", versus being injected with a poison, which would be considered murder. Of course, if they are on a ventilator, then that can be turned off as a quicker method of euthanasia. StuRat (talk) 02:09, 13 March 2010 (UTC)

There have from time to time, including recently, been reports in the UK press and media of, and legal investigations into, cases where elderly patients (some terminally ill, others apparently not) have allegedly been wrongly or over-prescribed with opiate drugs or denied nutrition and water, leading to their quicker and/or unnecessarily unpleasant and/or unnecessary deaths. One case is referenced here; I recall others I don't feel inclined to dig for. Such cases, obviously legally and emotionally sensitive, may reflect possible incompetence or wrongdoing on the part of some medical staff. 87.81.230.195 (talk) 04:23, 13 March 2010 (UTC)

The issue of over-prescribing opiates is often one of double effect - painkillers are given in the quantity necessary to deal with the pain despite the doctor knowing that dose will kill the patient. --Tango (talk) 21:22, 13 March 2010 (UTC)

So, if nutrition was not denied intravenously (say), would the patient live longer?--79.76.188.14 (talk) 22:49, 13 March 2010 (UTC)

Obviously the main factor here is how severe the cancer is, but without nutrition I imagine the patient would die sooner as they would be weaker from lack of energy. No source for that, just seems common sense. Regards, --—Cyclonenim | Chat  23:03, 13 March 2010 (UTC)

lab rats

where are lab rats and mice bought from for research purposes? —Preceding unsigned comment added by Thekiller35789 (talk • contribs) 23:36, 12 March 2010 (UTC)

From companies that sell rats and mice for research purposes. E.g., [2] See laboratory rat. alteripse (talk) 23:38, 12 March 2010 (UTC)

i already read that it didnt help —Preceding unsigned comment added by Thekiller35789 (talk • contribs) 03:52, 13 March 2010 (UTC)

here are several commercial suppliers: [3] [4] [5]. Note that most animal houses (individual labs rarely keep their own rats and mice these days - instead they are housed at a central specialised facility) would only buy animals if they needed a new strain or an existing strain was lost due to disease etc. Otherwise it is far easier and cheaper to breed their existing mice.131.111.185.68 (talk) 08:45, 13 March 2010 (UTC)

Also, while I haven't done work with animals myself (well, other than fruit flies) I would imagine that labs quite often give more specialised strains as gifts to other labs which need them. This is quite common practice in (academic) science - it is much less hassle than trying to agree on a price and the sharing of resources means that everyone can get on with their work without spending six months duplicating what another lab has already done.131.111.185.68 (talk) 08:54, 13 March 2010 (UTC)

galactic spiral arms

Given enough time will the stars that fill the galactic plane accrete and form "stellertoids" (similar to planetoids only made up of stars that have accreted into super giant stars or black holes) and become spaced out into planetoid like orbits around the galactic center? 71.100.11.118 (talk) 23:52, 12 March 2010 (UTC)

Not really. Depending on what the ultimate fate of the universe is, they might well all end up in black holes, but that will be long after they stop being fusing stars (they'll be neutron stars, white/black dwarfs, smaller black holes, etc.). I suppose it is possible for stars to merge, but it would be a very unusual event (and I'm not quite sure what would happen - it would depend on the masses of the stars involved, certainly). I think you may be a little confused about what a planetoid is, too - a planetoid is a small planet-like body. In the early solar system, planetoids would have accreted into planets. Given your terminology, I think you may be thinking planets merge together to form planetoids, which isn't the case (I apologise if I'm the one misunderstanding). Also, the orbits of stars around the galactic centre are very similar to the orbits of planets around stars - the spiral arms are just a kind of pressure wave caused by complicated interactions, individual stars follow normal orbits (they are slightly different because the mass of the galaxy isn't dominated by the centre like that of a solar system is, but that isn't very important, it just affects the numbers a little). --Tango (talk) 00:17, 13 March 2010 (UTC)

No, actually I am thinking of planetoids as the precursors to the formation of planets. Try to decide what terminology to use when composing an imaginary situation always runs the risk of bringing with it a misconception or the wrong idea (my naughty speller doesn't help either!). Basically what I'm asking is that in absence of any other interference such as collisions or merges with other galaxies or the Universe ending prematurely from the Big Rip or Crunch if accretion of stars into planet like orbits around the Galactic center is a reasonable consequence. (The main reason for the question is in consideration of how much space can separate such orbits since the motion of the solar system planets form spirals when you connect them as dots for regular periods of time.) —Preceding unsigned comment added by 71.100.11.118 (talk) 07:22, 13 March 2010 (UTC)

In that case, I guess the answer is "yes", but they won't be stars by the time it happens. They will be stellar remnants of various types and will merge into black holes. The same process will result in quite a lot of objects being thrown out of the galaxy altogether, though (unless something happens first, they will eventually fall into a black hole somewhere, anyway). --Tango (talk) 12:45, 14 March 2010 (UTC)

Also, before your scenario or something similar could occur, the Milky Way could likely collide with the Andromeda Galaxy to form Milkomeda. This would disrupt the spiral structure, and our galaxy could turn more elliptical. The colliding pair could also absorb numerous other galaxies such as the Large Magellanic Cloud, Small Magellanic Cloud and the Triangulum Galaxy, further disrupting the "clumps" of stars and gas in the galaxies. ~AH1^(TCU) 01:55, 13 March 2010 (UTC)

The merging of stars in the galactic disc is an excedingly rare event. Mergers are still rare but not completely negligible on higher density enviroments such as blobular clusters. Read blue stragglers to see what happens to stars after merging. Dauto (talk) 17:10, 13 March 2010 (UTC)

I assume Dauto has mistyped "blobular clusters" for Globular clusters, but I like this new term and can imagine someone applying it in the future, perhaps to non-symmetrical globulars :-). 87.81.230.195 (talk) 21:52, 13 March 2010 (UTC)

However, in the case of stars that collide with sufficient energy, they may explode in a hypernova, or if pulsars are involved then magnetic or gravitational waves may be released (I seem to recall an article involving a specific event where such a collision released energy that hit the earth's atmosphere but I can't find it right now). ~AH1^(TCU) 23:33, 13 March 2010 (UTC)