Wikipedia:Reference desk/Archives/Science/2009 June 3

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June 3

Earth is seriously moving away from sun?

Is earth seriously moving away from sun? I saw it was update like last night earth moving 20 centimeters apart each year they said because sun is losing solar winds and masses. Is it just earth or this is also happening to venus, mars, and others? If sun is losing mass now, then sun will just lose more and more, I don't think sun will gain mass back again.--69.229.240.187 (talk) 00:25, 3 June 2009 (UTC)

Yes. Tidal forces and loss of solar mass both result in the Earth receding slightly from the Sun. Without checking numbers, I'd expect tidal forces to be the dominant mechanism. The same effect occurs at varying rates on all other bodies orbiting the Sun. The Sun can gain mass (any time a long-period comet crashes into it, for instance) but is a net loser of mass. — Lomn 00:46, 3 June 2009 (UTC)

It is true, but it is a negligible amount. Our measurements of the distance from the Earth to the Sun are only accurate to a few metres (according to Astronomical unit), so a change of 20 centimetres isn't even measurable. It is happening to all planets, though. When the sun nears the end of its life it will throw off its outer layers, dramatically reducing its mass, and then the Earth will move significantly further away, but that won't happen for billions of years. --Tango (talk) 00:49, 3 June 2009 (UTC)

• Does all planets including Mercury and Venus? Venus have backwar orbit, so I always thought Venus will just end up like Triton vs.Neptune or Phobos vs. Mars. But how is Mercury moving farther when it is so close to sun?--69.229.240.187 (talk) 00:58, 3 June 2009 (UTC)

• I honestly can't dig anymore. I couldn't find the source said all the planets is now moving out. Formation and evolution of the solar system don't give us enough informations though.--69.229.240.187 (talk) 02:31, 3 June 2009 (UTC)

• • Venus does NOT have a backwards orbit. It has a backwards rotation, which means that it turns through a day in the opposite direction as the other planets do (or if you prefer, it is upside-down. Same difference). It turns so slowly, however, that a venusian "day" is longer than a venusian "year". And it orbits in the same direction as all of the other planets. However, as noted, the sun is losing mass via mass-energy conversion, which means that as the mass decreases, its gravity decreases over time. Since the force of gravity is decreasing, ALL objects gravitationally bound to the sun will drift farther away. It doesn't matter if we are talking about Mercury or the Oort cloud. Less mass = less gravity. Less gravity = larger orbits. Its that simple. --Jayron32.talk.contribs 03:02, 3 June 2009 (UTC)

Even if it was a retrograde orbit (which is not the case), it would still move out to a larger radius if the sun's mass decreased. Nimur (talk) 03:55, 3 June 2009 (UTC)

For perspective, at a rate of 20 cm / yr, the Earth's orbit would take 7.5 billion years to change 1% in radius. In other words, the current rate of expansion is utterly negligible for any practical purpose. Dragons flight (talk) 03:58, 3 June 2009 (UTC)

• If Dragon Flights is talking about in 7.5 billion years. It is still the same game as usual. Nothing have changed. Earth survival or not is still an even money (50/50). Mercury is going to go definitely. Venus' odd to doom is obviously better than a 50/50, Mars could be a doom one but highly unlikely. When sun expands, then it's surface area will increase drastically, then sun's mass and gravity will probably climb. This is possibly another story.--69.229.240.187 (talk) 04:26, 3 June 2009 (UTC)

Surface area expands OK, but why would mass/gravity climb? It will only decrease. - manya (talk) 04:54, 3 June 2009 (UTC)

We don't even know if earth will still exist at that time or not. It got to be some theory to make earth getting engulf, it's still a 50/50 odds.--69.229.240.187 (talk) 05:00, 3 June 2009 (UTC)

69': You have the ending of the sun all wrong. When it runs out of hydrogen and starts running on helium, it suddenly starts generating a LOT more energy (but not for very long) it blows off an immense amount of material - which is certainly enough to kill everything on earth, blow away it's atmosphere and oceans and generally trash the place. Once that's happened, both the mass and gravity of the sun are a lot less and the gravity can no longer stop the sun from expanding due to photon pressure from all the energy it's putting out - THAT'S why it expands. It doesn't gain mass or gravity - quite the opposite in fact!

You have to think of stars as fighting a continual battle between their immense gravity trying to collapse them into something much smaller and more exotic (like a black hole or a neutron star) and the photon pressure from the energy they put out that's trying to inflate them. If the gravity loses the battle, the star grows to a larger, dimmer red giant - if gravity wins, it shrinks to a white dwarf, a neutron star or a black hole depending on how much gravity there is. Ironically - the bigger the star is, the smaller it ends up being! The Sun - being relatively tiny gets bigger - but NOT heavier. SteveBaker (talk) 12:47, 3 June 2009 (UTC)

As the core fills up with heavier and heavier atoms it becomes denser and so so the pressure at the core increases (due to the stronger gravity) and the fusion rate increases. The sun expands because it is producing so much more energy. the expansion reduces the pressure in the core and slows down the rate of fusion. its a feedback process. (disclamer:I am not an expert) just-emery (talk) 20:04, 3 June 2009 (UTC)

I'm not getting the whole thing wrong just mass and gravity. I didn't know that in past, but I got it now. Yes, when sun gets bigger, it's mass and gravity plummets quickly, mass and gravity just dives way down. I got weight-size thing upside down. Yes I know neutron star and black hole comes from high mass star not sun since our sun is average mass star. yes, I know once sun runs out of hydrogen, earth ocean and life will run out first, earth will become a hell instantly until earth atmosphere runs out earth is a char garbage. But have you seen Formation and evolution of the Solar System#Future yet. They siad because of tidal interaction, earth will get destroy at end of sun's peak size, when mass and gravity drops suddenly. Studys 6 month ago hadn't change yet. Yes, if earth survives, then earth is a trash hell black-out scorchland with nothing on it. Maybe earth will be gone, destroy by sun. With tidal interaction between sun and earth I don't know what it is about.--69.229.240.187 (talk) 22:37, 3 June 2009 (UTC)

Actually what is a tidal force thing when sun gets to peak of it's diameter about? how can tidal force happen when gravity and mass drops.--69.229.240.187 (talk) 00:32, 4 June 2009 (UTC)

All the informations is on this site and source on Formation and evolution of the Solar System#Notes and references numbers 86-89, if my questions is clear fully.--69.229.240.187 (talk) 02:24, 4 June 2009 (UTC)

I haven't read all those posts yet, but the Earth really is moving away from the Sun, and it's not due to nuclear fusion: http://www.skyandtelescope.com/news/46618862.html. That article suggests tidal interactions between Earth and the Sun. Sky and Telescope is a reputable amateur astronomy magazine, but I don't know how scientifically reliable that specific article is. --Bowlhover (talk) 10:54, 4 June 2009 (UTC)

Main factor of earth moving away from sun is lost of mass.It is not only earth but all planets maybe moons vs. planets (Titan and Saturn) is also moving away so is Galilean moons vs Jupiter. Fron Section 15 they said because of tidal bulge. Earth is both moving farther and closer to sun. Is tidal bulge. Net solution is when sun gets old, hydrogen starts to deplete, then with less gravitational pull, then all planets and moons moves farther out. Tidal bulge is probably a friction, slows down earth's orbit, but I don't know how drag of sun and earth shrinking orbit works is my question.--69.229.240.187 (talk) 22:32, 4 June 2009 (UTC)

MM Experiment

When Michelson and Morley did their experiment they were looking for variations in the brightness of the light right? I was under the impresssions that the split beams were recombined into a eyepiece. As they rotated the apparatus they were looking for variations in the brightness of the light right? Because they knew they couldn't make the arms exactly the same length apart. They weren't looking for actual fringe shift, just the effect. —Preceding unsigned comment added by 24.171.145.63 (talk) 00:53, 3 June 2009 (UTC)

Basically you are correct. They could not measure the exact length of each arm with sufficient accuracy with any instrument at their disposal, but by looking at the interference pattern, you can obseerve changes in the difference between the lengths. Whe I set up an interformeter I used a laser, so I had a nice long coherence length to work with. With a laser, you start with two paths that ate roughly the same length. as you vary one length by one wavelength, you cah observe the pattern shift through an entire cycle. I used a mirror mounted to a piezoleletric actuator to vary the length through several wavelengths and observed the repeating cyclic pattern. When using a white light as Michelson did, you need to get the paths a lot closer to the same length to start with. -Arch dude (talk) 01:28, 3 June 2009 (UTC)

Just one more quick question:

http://spiff.rit.edu/classes/phys314/images/mm/mm3_rot.jpg On pg. 336 it says (little below the middle), that if the whole apparatus is turned through 90 degrees the fringe shift would double. Why? —Preceding unsigned comment added by 24.171.145.63 (talk) 04:34, 3 June 2009 (UTC)

That's what would happen if the speed of light *did* depend on motion through the "luminiferous aether". In actual fact, it doesn't, so there is no change. --Tango (talk) 14:56, 3 June 2009 (UTC)

Smallest number of neutrons

What is the smallest number of neutrons to start the formation of a Black Hole?

---- Taxa (talk) 01:04, 3 June 2009 (UTC)

Neutrons do not form black holes. Generally, they are formed by collapsed stars, as such they are called Stellar black holes (there are other types of black holes, but when most people talk about black holes, they mean stellar black holes). According to our article, which I link, there are several factors besides the mass of a star which will lead to it forming a black hole, but the threshold limit seems to be roughly 20 solar masses. That is, stars which are larger than 20x the size of our sun will generally form black holes upon their death, while stars smaller than that will not. --Jayron32.talk.contribs 02:57, 3 June 2009 (UTC)

Neutrons typically form neutron stars, which can not collapse to small enough size to form a black hole. This is because the Pauli Exclusion Principle forbids two neutrons from occupying the same spatial location with the same quantum state. This quantum-mechanical explanation can be "summarized": even though they have no electric charge, neutrons will repel each other if they get squished close enough together via a repulsive nuclear "force" (in truth, this is not a repulsive force because it can't be written as the gradient of an energy potential, which is why the quantum mechanical explanation is "better", but you can sort of conceptualize the idea). NeutronsPure neutrons can't get close enough together for gravity to dominate (which is required for black hole formation) - they must start off with other particle types present, or accrete extra mass in order to form a black hole. Nimur (talk) 04:02, 3 June 2009 (UTC)

Actually the degeneracy pressure can be expressed as a gradient of a potential in the mesoscopic limit. It's still not a force in the everyday understanding of the word as applied to isolated nucleons, but the potential formulation is plenty useful. For example, the Tolman–Oppenheimer–Volkoff limit describes the point at which degeneracy pressure is no longer able to counter gravity and a neutron star would inevitably become a black hole. According to the article it is between 1.5 and 3 solar masses (with uncertainties due to limited understanding of the behavior of nuclear matter at extreme density). So a neutron star can have at most something less than 3 solar masses, which translates to ~4×10⁵⁷ neutrons. Dragons flight (talk) 04:30, 3 June 2009 (UTC)

Of course all this applies for a neutron star under only gravitational contraction. If some other force compresses it, it should be able to go black earlier. Romulan force fields come to mind, but more realistically: What if two minimum mass neutron star smash into each other? Gamma ray burst? --Stephan Schulz (talk) 13:19, 3 June 2009 (UTC)

If you're asking how big a neutron star has to be before its gravitational force is too large for it to keep form collapsing into a black hole, I have no idea. If you're asking for the smallest number required for a black hole to be physically possible, regardless of what forces them together, the answer is a Planck mass divided by the mass of a proton which is 2.1764411×10^-8 kg / 1.672621637×10⁻²⁷ kg = 3.6403606×10¹⁹ neutrons. I think it's actually somewhat less than that, as the energy required to force the neutrons together would raise the mass. Also, the black hole would evaporate almost instantly. — DanielLC 21:45, 3 June 2009 (UTC)

Why does a black hole need to be at least the Planck mass? I believe a smaller black hole would have a Schwarschild radius of less than the Planck length, which is a somewhat meaningless concept, but does that mean the black hole can't exist, rather than just that our theory of black holes breaks down? --Tango (talk) 22:03, 3 June 2009 (UTC)

The latter. We don't have an accepted theory that can understand gravity at such small scales, so whether a black hole could have less than a Planck mass of material or be less than a Planck length in size is unknown. General relativity would no longer be an adequate theory to describe such objects. Dragons flight (talk) 08:28, 4 June 2009 (UTC)

But could you (hypothetically) take a single neutron and accelerate it until its mass exceeded the Planck mass, thus creating a black hole from one neutron (plus a lot of energy) ? Gandalf61 (talk) 15:59, 4 June 2009 (UTC)

The energy used to accelerate a particle does not help creating a black whole, after all there will always be a coordinate system in which the particle is at rest. Dauto (talk) 18:18, 4 June 2009 (UTC)

Okay, how about colliding two neutrons travelling in opposite directions at near light speed - maybe in a galaxy-sized LHC ? Could that (hypothetically) generate a high enough energy density to create a quantum black hole ? Gandalf61 (talk) 12:45, 5 June 2009 (UTC)

I think the problem lies in the definition of what a "black hole" is. A single point-sized particle (a neutron, for example) is a black hole according to some definitions. (Mathematically, there is some distance that lies outside the (zero) size of the particle at which it's gravity becomes so strong that light cannot escape.) Does that make it a black hole? Well, maybe, maybe not. You certainly can't get close enough to a neutron to be 'sucked in' as you would with a black hole created by a collapsing star. I think the problem here is not what nature does or does not do - but merely that we've decided to attach a name to a somewhat flawed definition. SteveBaker (talk) 17:26, 4 June 2009 (UTC)

If a neutron is a black whole, than the no hair theorem is not valid. Dauto (talk) 18:18, 4 June 2009 (UTC)

How to cure fungal plant pathogens

My Mango tree with leaf spots.

My Mango tree has some kind of fungus which is causing it's leafs to produce spots. I think it might be Cercospora capsici but I am not sure since there are many different kinds of plant fungus that might have similar outcomes.

I would like to know what I can do to get rid of the infestation and also how I can prevent it in the future. I have added a photo of the actual plant in question to better illustrate the problem. Joel M. (talk) 02:46, 3 June 2009 (UTC)

Sorry for the lack of responses - have you tried contacting a "master gardener"? Where I live, the local university seems to loan them out to public places like shopping malls and tree nurseries, where they run a (physical) reference desk of their own for a few hours a weekend. (Photo moved) Tempshill (talk) 20:30, 3 June 2009 (UTC)

Fungicide has a couple of natural things you can try that aren't that difficult to find. (If you can't find them at your local grocery store, health food store, drug store or pharmacy, try ordering online) I'm not sure how specific you'll have to target your fungus. Most fungicides seem to be pretty broad spectrum.71.236.26.74 (talk) 15:22, 4 June 2009 (UTC)

Understanding another part of the MM experiment

Scrap my old questions; I get them now. Anyway- I hope this doesn't go against your reference desk policy since this isn't homework that I'm not trying on. here on page 340, first line, it says something along the lines of the width of the fringes varied from 40-60 divisions (im interpreting it to mean there were between 40 and 60 fringes). It than says the average is 50 and says one division means.02 wavelength. Where did this last statement come from?

One divided by 50 is 0.02. They are taking the reciprocal. The "divisions" are not number of fringes - it's the width of each fringe, measured in "Divisions of the screw head" instead of "millimeters" or "centimeters". I don't know why they use such arbitrary units - it's just like the number of marks on a ruler or optical viewer or other measurement device, and needs some conversion to standard units. Nimur (talk) 15:55, 3 June 2009 (UTC)

Lens and mirror placed in water

We have been given following question in summer assignment:-
"A concave mirror and a convex lens are placed in water. What change, if any, do you expect in their focal length"
Since f = 1/R and curvature of lens and mirror won't change if placed in water, am I correct in assuming that focal length won't change? --shanu 05:36, 3 June 2009 (UTC)

For a reflective system, the main concern is the geometry of the surface; but for a refractive system, I think you need to consider the refractive index when calculating the focal length. DMacks (talk) 05:51, 3 June 2009 (UTC)

Take a look art the articles on mirror and refraction. A key point in the mirror article is that "a beam of light reflects off a mirror at an angle of reflection that is equal to its angle of incidence." A key point in the refraction article is "refraction occurs when light waves travel from a medium with a given refractive index to a medium with another." You may need to read more of the refraction article to better understand what this saying. Once you have a better understanding of the two principles, you can revisit the question of what effect water immersion has. A couple of related questions that may test your understanding: (1) When swimming, if you open your eyes underwater, can you see as well as above water? Why? (2) If you wear goggles or a face mask, does this change your vision? Why? (Hint: What shape is the outer surface of the face mask?). Feel free to ask follow-up questions after you've had a look. -- Tcncv (talk) 06:00, 3 June 2009 (UTC)

The question is discussed here. Cuddlyable3 (talk) 09:14, 3 June 2009 (UTC)

The key here is in the precise definition of refractive index. Since this is borderline homework - I'll point you to the "Definition" section of Refractive index - read it carefully. SteveBaker (talk) 12:35, 3 June 2009 (UTC)

Thanks for your help. By the way does it also mean that R = 2f won't work for lens in water? And also since thin lens formula will change, microscopes will be affected when placed in water.--shanu 05:21, 4 June 2009 (UTC) —Preceding unsigned comment added by Rohit..god does not exist (talk • contribs)

That should also help. Dauto (talk) 15:36, 3 June 2009 (UTC)

Thanks for your help. By the way does it also mean that R = 2f won't work for lens in water? And also since thin lens formula will change, microscopes will be affected when placed in water.--shanu 05:23, 4 June 2009 (UTC) —Preceding unsigned comment added by Rohit..god does not exist (talk • contribs)

For an importatn application, see Immersion lithography.-Arch dude (talk) 08:11, 4 June 2009 (UTC)

Sperm Count

Good day,

I was wondering - what is the time frame for the sperm count to reach its maximum ounce the person has ejaculated ? ( how long will it take from a count of 0 to reach 20 000 000)

Thank you

Marcello —Preceding unsigned comment added by Saladza (talk • contribs) 07:21, 3 June 2009 (UTC)

If I understand your question correctly, I do not believe it's normal for the sperm count to reach zero even after multiple ejaculations in a short space of time. If a sperm count is zero, that would likely indicate fertility problems like azoospermia or perhaps that the person has had a vasectomy. In fact according to our article "How long the man has abstained prior to providing the sample for analysis affects the results. Longer periods of abstinence correlate with poorer results - one study found that men with repeated normal results produced abnormal samples if they abstained for more than 10 days. It is recommended not to abstain for more than one or two days before providing the semen sample for analysis" although it's not clear if this includes sperm count or other factors like motility and morphology. Note that sperm counts are only really meaningful in when measuring them from ejaculations. Nil Einne (talk) 07:45, 3 June 2009 (UTC)

Three Step Process Of Lethal Injection

Why is it that we have a three step process for lethal injection, while pets only have one simple injection when they are getting put down? Wouldn't it be better just to use the one? It'd be cheaper and wouldn't last as long, and there have been no known accounts of pets still being alive and conscious yet paralysed when the final injection stops the heart and collapses the lungs (obviously because there is only one injection that kills them outright). Why not just do it like that? --KageTora - (영호 (影虎)) (talk) 10:52, 3 June 2009 (UTC)

You are evidently talking about the US procedure. There has been much debate about this - our article Lethal injection covers this in some detail. The idea is to use the first shot to briefly aneasthetise the victim - the second to relax the muscles so there is no embarrassing thrashing around - and the third to stop the heart. There is much controversy over this because there is a belief that the first injection might not work - or might not work for more than a couple of seconds - resulting in the victim being conscious but unable to move or speak (because of the muscle relaxant) while a painful heart attack ensues from the third injection...also the possibility of surviving the heart attack and then slowly suffocating while the muscle relaxant prevents breathing. For smaller animals (cats, dogs, etc) a single massive shot of barbiturates causes unconsciousness and then both heart stopping and a cessation of breathing in about 30 seconds. The problem is (as our Animal euthanasia article indicates), this doesn't work well on large animals because the barbiturate dose required is too high. I can only presume that humans fall into the "large animals" category...hence the controversial three-step process. SteveBaker (talk) 12:29, 3 June 2009 (UTC)

• Please avoid attempting a debate here by using loaded words like "victim". --Anonymous, 03:45 UTC, June 4, 2009.

Please avoid being uneducated in the use of the English language...pick up a dictionary sometime. From Wiktionary, meaning (2) for the word "victim" is: "Anyone who is physically harmed by another."...being killed by the state executioner certainly counts as being "physically harmed". I chose that word with great care. I you choose to pick a different, and perhaps more 'loaded' meaning - that's a debate of your own making! SteveBaker (talk) 15:39, 4 June 2009 (UTC)

To the talk page, please. --Anon, 08:00 UTC, June 5, 2009.

Also, a different ethical standard is clearly being applied to animal euthanasia than to human euthanasia (which is used more sparingly). It seems to follow that the procedures would reflect that difference in ethical concerns. Nimur (talk) 16:00, 3 June 2009 (UTC)

Having used barbiturates to anesthetize animals for surgery, I can tell you that they are pretty tricky. Even huge doses with small animals don't lead to quick death with 100% reliability -- once in a hundred times the animal continues to breathe at a very slow rate for quite a while. Barbiturates don't actually stop the heart, by the way -- so suppression of breathing is the way they kill. Looie496 (talk) 18:45, 3 June 2009 (UTC)

Animal euthanasia says there is cardiac arrest - is it incorrect? SteveBaker (talk) 20:03, 3 June 2009 (UTC)

Cardiac arrest is inevitable in a lethal injection, but barbiturates do not cause this directly, like potassium chloride does. cyclosarin (talk) 02:45, 8 June 2009 (UTC)

Pushing a helicopter

Suppose there's a helicopter (something small like a Bell 206, not a Chinook) hovering a few feet off the ground. Would it be possible for a human standing on the ground to push/pull the helicopter enough to move it, in any direction, without the aid of any other equipment? — Matt Eason ^{(Talk • Contribs)} 11:06, 3 June 2009 (UTC)

Well, there is no friction and very little air resistance involved - so all you've got to do is overcome the inertia - which is considerable. The classic Bell 212 helicopter weighs in at 3000 to 5000kg depending on fuel and passenger load (actually, technically, what we care about here is the mass not the weight - but it's still 3000 to 5000kg) so it would take quite a bit of effort to get it moving at any kind of speed - and quite a bit more to stop it: Force = Mass x Acceleration - so with a normal kind of force and a big mass, you don't get much acceleration. However even the smallest push would impart some acceleration and if you continue to push, the speed will gradually get faster and faster until you couldn't keep up with it anymore - so technically, in the purest theoretical sense - yes, you could push the helicopter around with no particular problem.

However, a helicopter doesn't just hover in one place passively. If the ground beneath has even a slight slope or unevenness - or there is any kind of wind - or a vertical surface is within maybe 50' in any direction - or if the helicopter isn't set up just perfectly - then without the pilot actively working to keep the helicopter still - it would drift off and spin all over the place at accelerations much greater than you could overcome with your puny muscles! So realistically, the pilot (or perhaps some autopilot hovering aid) is actively and continually making tiny adjustments to keep the helicopter still - and those adjustments would easily counter your puny efforts to move it. If he stops making the adjustments - then the helicopter is going to drift and you're not going to be able to stop it or make a significant difference.

So I'm pretty sure that the theoretical answer is "Yes" and the practical answer is at best "No" and at worst "Not a sufficiently precise question to yield a meaningful response"! SteveBaker (talk) 12:19, 3 June 2009 (UTC)

Very interesting - thanks Steve — Matt Eason ^{(Talk • Contribs)} 12:30, 3 June 2009 (UTC)

Here is a guy doing the fairly common stunt of pulling a train with his teeth. They claim 300 tonnes in this case, and the situation seems analogous to Steve's theoretical perfectly tuned helicopter, with far greater mass. --Sean 13:10, 3 June 2009 (UTC)

The practicality of this can be changed to a similar problem: Can the kick of shooting a gun move a helicopter? Humans are capable of countering the kick of large rifles. However, it was noted during Vietnam that firing rifles that were bolted down to the helicopter would cause it to rock enough to notice. So, humans can push with more force than the kick of the rifle and the kick of the rifle is enough force to cause the helicopter to rock. So, it is feasible that a human would have enough pushing force to cause a helicopter to noticeably move. -- kainaw™ 13:54, 3 June 2009 (UTC)

"Humans are capable of countering the kick of large rifles." But a human probably cannot counter the kick of a .50 caliber machine gun fired rapidly - it's sort of a totally different animal. Even small machine gun like a Squad automatic weapon requires a bipod or a tripod to keep it from blowing back out of control. A large gun such as a GAU-19 (standard fare on a Black Hawk) is going to impart tens or hundreds of kilogram-meters per second of momentum (according to our article, 500 pounds force sustained). I doubt any human sustain such a push. Nimur (talk) 16:06, 3 June 2009 (UTC)

There is no doubt about that. As TotoBaggins points out, humans can move trains that are far heavier than a helicopter. The only problem is the one Steve points out about the natural movement of the helicopter due to the difficulties of keeping it steady are going to be greater than the movement a human could produce. --Tango (talk) 15:37, 3 June 2009 (UTC)

You would probably be pushing below the center of mass, causing the helicopter to pitch or roll a little depending on where you push. I'm not sure which net effect this would have without corrections by the helicopter. PrimeHunter (talk) 16:04, 3 June 2009 (UTC)

OK - I'm going to have to explain helicopter aerodynamics...bear with me! A civilian helicopter - or a heavy-lift military one - would have rotors that bend upwards a little (it's called the "coneing angle"). If you think about the helicoptors rotors - instead of being a couple of separate blades that are sweeping around really fast - imagine them as if they made a solid disk...that's kinda what it looks like. So that when the weight of the helicopter's body is supported by the rotors - the rotor disk is pulled down into a cone (with the point of the cone pointing downwards). Now, when the helicopter rolls (or pitches) that tilts that cone - right? So one side of the rotor disk/cone is now more horizontal and the other side is more steeply tilted upwards. As the rotors travel around the surface of the cone - the one that's moving more horizontally to the ground pushes down the air directly towards the ground - getting the maximum possible upward push - but the rotor that's on the opposite side is nowhere near horizontal - its at some angle to the ground - so the air it pushes goes out at an angle. This does two things. Firstly, it means that when you push on the skids at the bottom of the helicopter - the steeper blade (which is on the far side of the helicopter) pushes the air slightly away from you - which makes the helicopter try to push back against you. But because that air isn't being pushed as hard against the ground as on the blade that's nearest to you - the far side of the helicopter loses a bit of lift - and the side that's nearest to you gains a bit...for as long as you push. This tends to make the helicopter level back out again. So the effect of this coning angle thing - is to give the aircraft some inherent stability...if you make it roll - it tries to roll back to the level position again. That's actually the only thing that makes the helicopter flyable...it would be impossible to have good enough reaction time if it didn't do that. So the fact that you are pushing below the center of gravity turns out not to matter very much! The other thing that helps with stability is that the center of gravity is about halfway up the helicopter - where the engine, gearbox and fueltank probably are. However, the lift from the rotors pulls upwards from the top of the 'mast' - so (in physics 101 terms) you have a 'moment' - you have gravity acting on the center of gravity and the lift acting on the top of the mast. So long as those two forces are in a straight line - nothing much happens. But if the helicopter rolls (or pitches) the two forces are no longer in a straight line - which imparts a rotation - which in turn tries to keep the helicopter level. It's as if the body of the chopper was a plumb-bob hanging under the rotor disk. Anyway - I didn't want to complicate the earlier discussion with all of this complicated stuff. Suffice to say - this ISN'T the reason you couldn't (in theory) push a hovering helicopter. SteveBaker (talk) 19:58, 3 June 2009 (UTC)

Is it possible that the "pushing back against you" bit you mention would be strong enough to actually result in the helicopter moving towards you when you pushed it away from you? That would be a rather interesting bit of trivia. --Tango (talk) 22:00, 3 June 2009 (UTC)

I don't see how. But helicopters are complicated machines - all sorts of bizarre gyroscopic effects - and the way the tail-rotor figures into things...it's tough to reason all of the forces out. SteveBaker (talk) 22:50, 3 June 2009 (UTC)

It's irrelevant to the original question, but it might be worth noting that the effect Steve describes in relation to coning has an analog on some airplanes. Rather than both wings being in a single plane, they may be tilted slightly to form a gentle V-shape, called "dihedral", and this contributes to the airplane's stability in the same way that Steve describes. On the other hand, on fighter aircraft where a bit of instability is desired so the plane is more maneuverable, the tilt may be reversed, which is called "anhedral". See dihedral (aircraft).

One other point. Steve mentioned the helicopter pushing the air "against the ground". Either an airplane or a helicopter makes its lift by pushing air toward the ground, i.e. down. Newton's third law and all that. However, if the vehicle is near enough the ground that the air is pushed substantially against the ground, it gets more lift, as the ground bounces it back up: see ground effect in aircraft.

--Anonymous, 04:02 UTC, June 4, 2009.

"… the wing keeps the airplane up by pushing the air down" eh? True but useless. An aerodynamic force on a body moving through a fluid is accompanied by an equal and opposite force on the fluid, but one does not "make" the other. A body which pushes air down generates lift—a body which generates lift pushes air down.—eric 06:43, 4 June 2009 (UTC)

(Actually - I said it pushes the air "towards" the ground - not "against" it - I picked my words carefully.) I assume that if you're pushing against a helicopter - that it's close enough to the ground for "ground effect" to matter. That greatly increases the amount of lift the helicopter has - but it doesn't change the 'coning angle' effect - which is indeed similar to the effect of dihedral on a fixed-wing aircraft - except that in the case of the helicopter it helps pitch stability as well as roll.

Eric's criticism is technically valid but I can't agree that the original statement was useless - it enabled both he and I (and hopefully, the OP) to perfectly well understand what was being described...and I think it's a clearer statement to the layman. We have to tailor our ref desk responses to a typical layperson - because we can't make assumptions about their scientific background. Saying that pushing the air down keeps the plane up is a perfectly valid statement - it merely fails to explain the REASON why pushing air down keeps the plane up. But then we also say that gasoline propels your car along the road without going into the details of the fluid dynamics, chemistry, thermodynamics and mechanics that makes that happen. Nit picking is unnecessary here - so long as the meaning is clear. 15:32, 4 June 2009 (UTC)

The lice were in Anonymous' hair, not yours. You were explaining dihedral, he was directly discussing lift, and based on his response i suspected did not fully understand your shorthand. Apologies if this was an incorrect assumption.—eric 16:23, 4 June 2009 (UTC)

I think there's been some misreading here. I was the one that mentioned dihedral (for airplanes); Steve was talking about coning (of a helicopter rotor) and I pointed out, as a point of interest, that the principle can extend to airplanes. As to the other point about "against the ground", if Steve would kindly reexamine his carefully worded posting, he will find the phrase in it. --Anon, 17:17 UTC, June 4, 2009.

Monitoring blood glucose levels

A friend of mine is diabetic and she usually tests herself once before a meal and two hours afterwards. I was curious why she has to wait two hours. After a meal, does blood sugar slowly climb to its maximum point after two hours? Or does it quickly surge high and then comes back to a "normal" level over a two hour period. Why not test one hour after eating? or three? --68.92.139.62 (talk) 12:38, 3 June 2009 (UTC)

It depends to some extent on what she's eating. If it is high in sugar then it will increase her blood sugar level pretty quickly. If it more complex carbohydrates, or mostly protein, say, then it takes longer to digest. You may find glycemic index interesting. --Tango (talk) 12:51, 3 June 2009 (UTC)

The timing of glucose testing in diabetes management mostly has to do with the medication regimen the patient is taking. In patients who take insulin, there are different formulations that have short-term (i.e. within a couple hours) and long-term (i.e. over the course of 24 hours) effects. See insulin therapy for details. Every patient will have a different regimen, depending on their own situation. One typical regimen is to take long-term insulin to maintain blood sugar throughout the day, coupled with short-term insulin doses corresponding to each meal to cope with the influx of glucose that happens after eating. The "fasting" blood test (before the meal) is meant to verify that the long-term insulin dose is correct. The 2-hour "post-prandial" blood test checks that the insulin that was taken with the meal was appropriate. You can see from the glycemic index article that in normal individuals, the blood glucose should be about back to normal by 120 minutes = 2 hours. This is the result of the pancreas releasing a burst of insulin, which is what is being simulated by the dose of insulin at mealtime. This, along with the usual time of action of the short-term insulin, is why 2 hours is a good time for a post-prandial check. Often, the patient will be given instructions about what to do if the post-prandial glucose level is too high (take some more insulin) or too low (eat something). The doctor managing the diabetes treatment will look at the test records to make sure that the dosage regimen is appropriate. --- Medical geneticist (talk) 13:33, 3 June 2009 (UTC)

Very interesting. Now my friend doesn't take insulin. It's all diet controlled. --70.167.58.6 (talk) 23:17, 4 June 2009 (UTC)

Energy of a point charge

I have read that that the self energy of a point charge, that is, the energy required to assemble a point charge, is infinite, and i am still struggling to come to terms with it. Does it mean that there cannot exist any point charges in the universe, or does it mean that it is just an embarrassing result of classical electrodynamics? I mean, you can say even the fundamental units of charge, the electron and the proton, are not point charges, but i am asking in principle. Also, we have found that even these fundamental entities are made of quarks, and quarks are made up of what not i don't know, but is there a limit? Say we somehow find the structure of a quark, and find its made up of little xions, and now we start to analyze these xions-its back to square one... is there a limit to exploring the structure of matter? And going by this notion that there can be no point charges, will we ever be able to find an end to this non stop search inside an atom ? Rkr1991 (talk) 12:52, 3 June 2009 (UTC)

The energy required to reduce the separation between two like charges increases as they get closer together. Roughly speaking, the energy for the last bit involves dividing by zero, which gives you infinity (the more precise answer involves improper integrals). We often think about fundamental particles as being point particles, but that's really just because we don't really know what they are, talking about sizes at that scale is largely meaningless because of quantum mechanical effects. So we have to make exceptions for fundamental particles, but any charge made up of more than one such particle can't have zero size. --Tango (talk) 14:39, 3 June 2009 (UTC)

You can't get around this problem by supposing that point charges exist a priori, because even an always-present point charge has an electric field, and the energy in the field (½ ∫ E²) equals the energy required to assemble the point charge from infinity. The field energy shows up as inertial and gravitational mass of the particle. This is a classical result, but in quantum field theory the problem gets worse, not better. The workaround is renormalization, which amounts to treating the particles as though they had a small but nonzero size, or introducing some other small-scale cutoff with a similar effect. Fortunately the predictions of the Standard Model are independent of the cutoff scale as long as it's small enough (the Standard Model is renormalizable). This is understood to mean that the Standard Model is just a large-scale approximation to the real physics, which presumably avoids the infinity in some unknown way, possibly by being discrete or (as in string theory) by distributing the charge over a 1-D structure instead of concentrating it in a point. This is a lot like the use of calculus to model systems that we know are discrete, like fluids or biological populations. -- BenRG (talk) 16:00, 3 June 2009 (UTC)

A finite amount of charge in a one dimensional string would also have infinite amount of energy in its field unless it was infinitely long. the idea that the mass of an electron comes from its electric field through self induction has long be abandoned. just-emery (talk) 20:12, 3 June 2009 (UTC)

BenRG, sure you must be aware that what you described (the introduction of a small arbitrary size) is not the renormalization itself but an essential housekeeping step - the regularization - that must be taken before the renormalization proper can be performed. There are several different ways to regularize the formally divergent integrals that show up in the solutions. Not all of those regularization schemes are based in the introduction of a cutoff. The renormalization itself is done by carefully subtracting the physically unorbservable (and often formally divergent) part of those integrals and keeping only the physically relevant terms. All point particles self energies gets taken care of that way. Dauto (talk) 21:25, 3 June 2009 (UTC)

Well i am able to understand and few points but unable to certain others (having had no formal education in QM), so i can only ask where this leaves us. Does it mean we shouldn't ask questions like what is the energy of a point charge, or that QM has somehow overcome this problem and say this much Joules is the energy of the point charge, or that we are still in the dark and don't know what to make of things ? Can there exeist any point charges in the universe, or is that forbidden ?Rkr1991 (talk) 05:00, 4 June 2009 (UTC)

A little bit of each. Firt: yes, you probabily shouldn't be asking that question because a point particle is a theoretical construct any ways which may not be a valid description of nature, second: yes, QFT solves the problem but it does so by sweeping it under the rug, and third: yes, we are somewhat still in the dark since we don't have a final theory yet. Dauto (talk) 17:52, 4 June 2009 (UTC)

Is it possible to set your own innards on fire?

Is it really possible to set your own innards on fire if you're chainsmoking while drinking your Neutral grain spirit (Everclear and similar) straight? --90.240.197.75 (talk) 14:46, 3 June 2009 (UTC)

The autoignition temperature of ethanol is 425°C ([1]). I doubt the smoke from a cigarette could get it up to that temperature, although I can't find a source for the temperature of cigarette smoke... --Tango (talk) 15:34, 3 June 2009 (UTC)

I'm trying to imagine where the oxygen to maintain the combustion would come from. I guess you'd have to keep taking deep breaths, but then ... Richard Avery (talk) 15:56, 3 June 2009 (UTC)

Did you swallow the booze and then swallow the cig, otherwise I can't imagine smoking gets temps that high inside a person. I am skeptical. 65.121.141.34 (talk) 16:04, 3 June 2009 (UTC)

It may be feasible to, with a combination of a lit cigarette and alcohol vapors, set your face and/or inside of your mouth on fire. But there is no way I could imagine that any such combination could burn your internal organs like lungs or GI-tract. --Jayron32.talk.contribs 18:13, 3 June 2009 (UTC)

I could sorta kinda imagine your breath catching fire (although it seems unlikely) - but not your throat or stomach - there is just not enough oxygen down there to sustain anything like that. SteveBaker (talk) 19:40, 3 June 2009 (UTC)

Sure, an accelerant and central nervous system depressant (Everclear), ignition source (cigarette), make it an obese person who maybe does not wash their clothes very often, and you've all the makings of a case of spontaneous human combustion.—eric 23:30, 3 June 2009 (UTC)

Was anyone else reminded of Helpless (Buffy the Vampire Slayer)? —Tamfang (talk) 18:49, 4 June 2009 (UTC)

What is VCR doing to my TV signal?

I have a coaxial cable from the aerial going into the VCR input socket, and another short coaxial cable connecting from the VCR output socket to the TV. I've noticed that when the VCR is completely unplugged from the power, rather than being just on standby, then the TV signal almost disapears with a very bad very noisy picture quality. So my question, please, is what is the VCR doing to my TV signal? I thought the input and output sockets on the VCR were just passively connected, but apparantly not. 89.243.113.64 (talk) 20:36, 3 June 2009 (UTC)

Almost exactly the same question was asked further up: Wikipedia:Reference_desk/Science#Poor_TV_picture_improves_a_lot_when_VCR_on. See if that answer is any help. --Tango (talk) 20:51, 3 June 2009 (UTC)

I've already seen that thanks. This is a different question about the same items. 89.243.74.161 (talk) 08:53, 4 June 2009 (UTC)

If the only connection between the VCR and the TV is a coaxial cable, that single cable carries either the off-air signal from the aerial OR the video from the VCR in the form of a modulated r.f. signal. The choice is made by an active switch circuit in the VCR. Without power the switch circuit can't pass either signal. Cuddlyable3 (talk) 10:13, 4 June 2009 (UTC)

The VCRs I've owned have passed the input coaxial signal to the output coax when powered off. -- Coneslayer (talk) 17:16, 4 June 2009 (UTC)

You're right the question is largely unrelated but as mentioned in the above question, VCRs do amplify the antenna signal. This is done I believe because otherwise the VCR would antenuate the signal, as it is using it. This amplification is active as long as the VCR is on, regardless of whether it is in standby or fully on. As mentioned above, most VCRs are also capable of adding a additional signal on a selectable channel to the feed so that you can receive the VCR on your TV if your TV has no other inputs. Nil Einne (talk) 00:03, 5 June 2009 (UTC)

I would expect the input to output of a VCR to include a powered buffer amplifier. (The phrase "common collector" comes to mind, or some reason). Edison (talk) 04:57, 5 June 2009 (UTC)

Is there any connection?

Is there any correlation between people who have Irritable Bowel Syndrome and Panic Attacks? —Preceding unsigned comment added by 86.167.247.150 (talk) 21:23, 3 June 2009 (UTC)

I am not a medical practitioner and this is merely an observation from articles I read on Wikipedia. I seems that selective serotonin re-uptake inhibitors are listed as possible treatments for both conditions. I can imagine how frequent panic attacks could result from low serotonin levels and the same could possibly be true for IBS. If you added obsessive compulsive disorder to that list then it would definitely have my vote. But again, I'm not an expert or a professional and you should probably ignore me.

It's unlikely that you'd get an official answer here as that would violate the terms of this page. 196.210.200.167 (talk) 16:32, 4 June 2009 (UTC) Eon

A brief search of medical journals suggests that there is definitely a statistical correlation between IBS and panic disorder. Additionally, the two occur more frequently in females, have similar age of onset and can be precipitated by stress. People with IBS have a higher prevalence of several psychiatric conditions and people with PD may also experience IBS-like symptoms during attacks. Finally, both seem to respond to similar treatment. - cyclosarin (talk) 08:12, 7 June 2009 (UTC)

Space inside a Black Hole

It appears then there are other requirements to form a Black Hole besides minimum number of neutrons. Is there a complete and ordered list of requirements, in addition to minimum number of neutrons, necessary to form a Black Hole? Also can a Black Hole be described (or defined) as matter in space which contains no space? ---- Taxa (talk) 21:57, 3 June 2009 (UTC)

There is only one requirement for a black hole - that the matter in question be contained within a ball of radius smaller than the Schwarzschild radius corresponding to the mass of the matter. (Actually, that's for a non-charged, non-rotating black hole. Without those assumptions you need a slightly more complicated formula, but the principle is the same.) --Tango (talk) 23:19, 3 June 2009 (UTC)

Applications of invisible light frequencies

I need a summer project for school. I have understood that variable lighting is a problem in computer vision, my just physics is not quite strong enough to tell if there are light frequecies that are not so dependent on sun's position on the sky or nearby lamps as visible light and cheap to generate/capture in good quality. I have feeling the answer is no, but it would be better to know for sure. Anyone have any idea? --194.197.235.28 (talk) 23:02, 3 June 2009 (UTC)

Actually this is quite commonly used in image processing. For example, a lot of toll-booths have a camera to catch the license plate of anyone who does not pay the toll. To cope with varying light conditions, the cameras are often sensitive only to infrared light, and the area is illuminated by an infrared bulb. This reduces interference from other sources of light and provides a controlled, constant illumination for the Optical Character Recognition program which will identify the vehicle. One reason this method is easy is because a lot of digital CCD cameras are already sensitive to infrared, so visible light can be filtered out (or left in as supplemental illumination). You might find Infrared photography interesting as well. We also have Thermographic camera, which describes passive infrared photography (using infrared cameras without an illuminating infrared lightbulb). These are commonly used as a type of night vision (not to be confused with low-light amplification). Thermographic infrared images make it very easy to spot vehicles, trucks, humans, and other hot objects, and are also often used in automatic image-processing (for example, automatic aim correction on a combat helicopter). Nimur (talk) 02:41, 4 June 2009 (UTC)

It's important to note that, while both called "infra-red", the light used by thermographic cameras and the light used by standard IR night vision cameras is very different. The latter uses "near infra-red", that is, light with wavelengths just slightly longer than that of visible red light. Thermal cameras use more distant infra-red. IR spans about 3 orders of magnitude compared to less than one spanned by visible light. There is a far greater difference between near and far IR than between violet and red visible light. If it were up to me, I would give near and far infra-red different names (with the boundary corresponding to 100°C). --Tango (talk) 02:21, 5 June 2009 (UTC)

It's not the frequency of the light - after all, the computer's camera only really sees red, green and blue - and some computer vision is done in black and white just because it's easier and the color doesn't really help much. No - the problem is uneven lighting - where some things in the scene are lit by brighter light than others - or when the lights flicker or change brightness while the computer is gathering imagery - or when the light is strongly directional producing super-bright highlights and deep shadows. You really start to appreciate how amazingly adaptable human vision is. SteveBaker (talk) 02:43, 4 June 2009 (UTC)

Right - the point of using active infrared illumination is to control the illumination environment. We could probably also control the visible-light environment too - but in the case of covert military or traffic cameras, flashing visible light around the imaging target is not an option. Using "invisible" infrared allows the operator to produce uniform illumination for the computer-vision system, without directly affecting the human-perceptible parts of the environment. Also, sometimes infrared just has better signal-to-noise qualities - I used a (mostly) infrared beacon for my robotic target-tracker project last year. Nimur (talk) 02:49, 4 June 2009 (UTC)

That answered my question perfectly. Thank you. --194.197.235.28 (talk) 15:17, 4 June 2009 (UTC)