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October 14

difference between quantum state and wavefunction?

Hi, what is the difference between a quantum state and a wavefunction? Can you change one without changing the other? Does every particle or system have both a quantum state and a wavefunction, even, say, completely static ones like ground state electrons?? Thanks in advance, It's been emotional (talk) 00:08, 14 October 2010 (UTC)[reply]

The quantum state is the set of quantum numbers that completely define a unique quantum system, for example, a single electron energy state, with spin state, in an atom. The wave function is the function which describes the probability space where a particle can be found. In other words, the wave function describes the shape and density of the volume of space where an particle (like an electron) is likely to be found. The three-dimensional plot of a wave function of an electron, for example, describes the shape of the orbital that contains that electron. Depending on your perspective, you can think of a wave function as either 1) a "scatter plot" graph which describes the location of an electron in the space around an atom as measured over an arbitrary length of time or 2) as a mathematical description of the shape of a standing wave centered on the nucleus. Perspective 1) relates to the "particle half" of an electron's nature, while perspective 2) relates to the "wave half" of an electron's nature. Quantum physicists don't make a distinction between these two perspectives, because electrons don't either, but you can think of either definition as being what a "wave function" is. So, here is how quantum state and wave function are related. The wave function is the mathematical equation itself which describes this space, while the quantum numbers are the variables in that mathematical equation. The distinction between the two is kinda like the distinction between a mathematical equation (x² + y² = r²) which will graph a circle, and the english definition of a circle as "the sum of all points equidistant from a fixed point". They both describe a circle, but they are not identical definitions. --Jayron32 01:48, 14 October 2010 (UTC)[reply]

Thanks very much for that answer. It clarifies things a lot. I have one further question, though: you seem at the beginning of your answer to be suggesting the wavefunction only represents the point in space where an electron is located. Does the wavefunction also represent the electron's spin quantum number? In other words, should I go more by your final point that a wavefunction and a quantum state describe the same information in different ways (as in the mathematical and the English descriptions of a circle), or is there any exception to this summary? It's been emotional (talk) 17:33, 14 October 2010 (UTC)[reply]

No, the wavefunction also contains the spin quantum number as a variable, I am pretty sure the wavefunctions of two electrons which differ only by their spin quantum number occupy the same space, but are orthogonal in some manner; this is why two electrons can occupy the same orbital but with opposite spins; their wavefunctions are perfectly orthogonal so that they don't interact with one another in any way (either constructively or destructively). Admitedly, my quantum chemistry is about 12-13 years sketchy at this point, so I forget exactly how they are orthogonal, but it's probably "orthogonal" in the same way that electrons have "spin", i.e. electrons don't have actual spin; but the property of particle spin behaves mathematically like angular momentum. In the same manner, the wavefunctions of the spin-opposite particles would be orthogonal, not in the sense that they are at physical right angles to each other, but in the sense that orthogonal waves do not interact. I wish I could explain that better, but as I said I am a bit rusty. Someone who remembers more than me will be along shortly to set it straight. --Jayron32 22:32, 14 October 2010 (UTC)[reply]

They're orthogonal in the math sense. That is, their inner product is zero. When discussing the standard inner product on vectors in 3D, the inner product being zero is equivalent to the vectors being at right angles (assuming they're non-zero) which is why this property it called "orthogonality". However wavefunctions don't live in Euclidean 3 space but some abstract infinite dimensional Hilbert space of all possible states. Rckrone (talk) 22:57, 14 October 2010 (UTC)[reply]

Great answer, thanks to you both. Glad I pursued it. It's been emotional (talk) 23:17, 14 October 2010 (UTC)[reply]

Contact lens solutions don't drip out of bottles when inverted.

Ok, they do, but the flows always stop. Is that atmospheric pressure at work? 67.243.7.240 (talk) 00:12, 14 October 2010 (UTC)[reply]

Yes, atmospheric pressure plays a large part. Initially, (with an open aperture at the tip of the bottle) the pressures inside and outside the bottle would be equal, and liquid will simply respond to gravity by flowing downward; however, if the bottle is un-vented (which would be true in this case) then this downward movement of the liquid lowers the pressure in the bottle, such that the greater atmospheric pressure (compared to the pressure in the bottle) exerts a force greater than gravity. If not for surface tension, the drops might still flow downward, replaced with upward (into the bottle) flow of air, but the small aperture plus surface tension are enough to keep the drops from leaking out. -- Scray (talk) 01:20, 14 October 2010 (UTC)[reply]

More drops will drip from a flexible plastic bottle than a rigid glass or metal one. Cuddlyable3 (talk) 12:55, 14 October 2010 (UTC)[reply]

Or maybe they have one of those plastic tubes connecting the tip of the bottle with the bottom? – b_jonas 13:04, 17 October 2010 (UTC)[reply]

Space telescope naming question--Edwin Hubble got one, how about Henrietta Leavitt?

Hi, does anyone know how the names for various NASA et al telescopes are decided on? Has Leavitt's name been bandied about? Thanks, Rich Peterson199.33.32.40 (talk) 00:39, 14 October 2010 (UTC)[reply]

I would expect the method varies a lot depending on the particular case. For one example Fermi Gamma-ray Space Telescope#GLAST renamed FGST. There's almost definitely a lot of politics involved. Nil Einne (talk) 04:34, 14 October 2010 (UTC)[reply]

When visiting Houston a couple of years ago some staff told me that there is a lot of politics involved in deciding on names, i.e. no formal process. HiLo48 (talk) 06:39, 14 October 2010 (UTC)[reply]

I see. Thanks to you both. Rich (talk) 06:47, 14 October 2010 (UTC)[reply]

Some space probes are named by contest. The Mars Exploration Rovers were named by a essay contest, as was The earlier Mars rover (Interestingly, pathfinder was named the "Carl Sagan Memorial Station" after the mission ended, in what I assume was an internal decision). In both cases there's still obviously a bit of subjectivity and politics that occurs in choosing the winner, but at least they're getting input from the public. The Mars Science Laboratory's name "Curiosity" was chosen by an online poll.

Some of the names have pretty whimsical sources. The telescope and UV spetrograph on the New Horizons probe are named, respectively, Ralph and Alice. Alice is named for the earlier instrument on the Rosetta (spacecraft), while Ralph is named for Alice's husband in The Honeymooners. It's unclear how Rosetta's UV spectrograph came to be known as Alice. Rosetta itself appears to be named after the Rosetta Stone [1](I guess this will help people understand comets the same way the Rosetta stone helped people understand Egyptian Hieroglyphics), and the landing probe is appropriately named the Philae lander, after a place where an inscription was found that helped refine our understanding of Hieroglyphics. It would be interesting to know the process by which those two names were chosen, but I don't see anything off hand.

Unsurprisingly, many names of satellites and instruments are acronyms. Sometimes, though, an acronym is forced to get a word that's somehow related to the mission; STEREO is a pretty good example of this. Buddy431 (talk) 14:42, 14 October 2010 (UTC)[reply]

See also Acronym and initialism#Contrived acronyms. Here are a few other suspects: AGILE, APEX, CHIPS, INTEGRAL, QUIET, WISE. PrimeHunter (talk) 23:05, 15 October 2010 (UTC)[reply]

Synthesis of 3-Carbethoxycoumarin

What is the synthesis mechanism for 3-Carbethoxycoumarin? Your article on Salicylaldehyde doesn't provide any detail. —Preceding unsigned comment added by Milt50 (talk • contribs) 01:40, 14 October 2010 (UTC)[reply]

I added a key detail to the item in salicylaldehyde. DMacks (talk) 04:40, 14 October 2010 (UTC)[reply]

magnitude of contact forces

two boxes, one on top of the other in an elevator. The bottom block weighs 30 N, the top weighs 20 N. The combined weight feels like 70 N. What is the magnitude of the contact force between the two boxes?

So I've solved that the total mass is 5.10 kg (3.06 kg for the bottom box and 2.04 for the top box). The acceleration of the elevator must be 3.92 m/s^2 upward to account for the difference in weight. I'm completely confused on the magnitude of the contact force though--i know that the force between the boxes should be equal in magnitude but point in opposite directions. Help!24.63.107.0 (talk) 05:12, 14 October 2010 (UTC)[reply]

Use Newton's 2nd law to find the net force acting on the top box. That net force must be the difference between the contact force and the weight of the top box. 174.58.107.143 (talk) 06:02, 14 October 2010 (UTC)[reply]

(edit conflict) Well you've done the hard bit! Now just forget about the lower box: imagine that the top surface of the 3 kg box is actually the "floor", and calculate the apparent weight of the 2 kg box. That is the contact force between the two boxes. Physchim62 (talk) 06:05, 14 October 2010 (UTC)[reply]

You have been told the contact force between the bottom box and the floor of the elevator is 70N. You only have to determine the contact force between the top box and the bottom box. Dolphin (t) 06:15, 14 October 2010 (UTC)[reply]

OP here, If I understand Physchim correctly, then I merely multiply 2.04 kg by the 13.72 (force of gravity + acceleration of elevator) to get a contact force of 28 N? Sweet!24.63.107.0 (talk) 06:20, 14 October 2010 (UTC)[reply]

Yup! Except it's even easier than that... For a bonus point, can you see a way to resolve this problem without solving for the mass of the boxes? Physchim62 (talk) 09:01, 14 October 2010 (UTC)[reply]

Weight x WeightAmplification = 20 x 70/(30 + 20) = 28 N Quod erat demonstrandum. Cuddlyable3 (talk) 12:51, 14 October 2010 (UTC)[reply]

What Cuddlyable3 is calling WeightAmplification is also known as load factor. Dolphin (t) 21:46, 14 October 2010 (UTC)[reply]

Yes, it gets a bit confusing to talk about different weights, but that's the terminology that the OP used. Cuddlyable's got the method, though. Another way to look at it is to consider the opposing contact force, the one in the upwards direction. There has to be one component of this force to counterbalance the weight of the box (due to gravity) and another component to accelerate the box upwards. We're told that the total contact force at the floor of the elevator is 70 N, and that the weight of the boxes due to gravity is 50 N. Hence there is a 20 N force accelerating the boxes upwards (with the rest of the elevator, obviously), so the acceleration must be 2⁄5 g. Physchim62 (talk) 02:20, 15 October 2010 (UTC)[reply]

Walking or Cycling?

As a form of exercise, which is more efficient, walking or cycling, in terms of energy expended against distance covered? Thanks asyndeton talk 08:38, 14 October 2010 (UTC)[reply]

Interesting phrasing. Walking burns more energy per mile, so I suppose in that sense it's "more efficient" as a form of exercise, assuming that for some reason you wish to cover as few miles as possible. --Trovatore (talk) 08:42, 14 October 2010 (UTC)[reply]

Of course a lot depends on how vigorously each exercise is performed, a 1km cycle sprint would expend more energy than a leisurely 1km stroll. -- Q Chris (talk) 08:49, 14 October 2010 (UTC)[reply]

(ec) Actually, I doubt that. A lot more per unit time, sure. But it would take maybe 10% to 20% of the time. It gets complicated if you start trying to figure in increased metabolic rate after the exercise, especially since how much that happens depends a lot on how much exercise you do regularly. --Trovatore (talk) 08:57, 14 October 2010 (UTC)[reply]

From a physics viewpoint, walking the 1 km should expend more energy since walking is less efficient (physics definition) than cycling, but Q Chris is probably correct from a physiological viewpoint because the sudden burst of vigorous effort resets the metabolic rate. "Efficiency" is a confusing concept in this context. I would suggest that sprinting the 1 km would burn more energy in total than either walking or cycling because it requires a vigorous effort and is intermediate in (physics) efficiency between walking and cycling. I'm not sure where jogging comes in the efficiency scale. Dbfirs 08:56, 14 October 2010 (UTC)[reply]

I'm somewhat with Trovatore on this one in being unsure. Definitely if you compare cycle sprinting vs leasurely strolling for the same amount of time, the cycle sprint will expend more energy. In this case I think it's far less clear cut and likely to depend on other factors. Nil Einne (talk) 10:41, 14 October 2010 (UTC)[reply]

Are not people confusing energy with power? Cycling 1km will require less energy than walking 1km, but if you sprint-cycle then you would use more power than walking, but the duration would be much less, hence the total energy should be less. The answer to the original question is, cycling would be more efficient as a form of transport. Cycling requires less energy than walking to travel the same distance. I understand that this is because when walking you use energy to stay upright, and much of this is saved when sitting down on a bicycle. In terms of being an "efficient" means of exercise, than whatever gets your heart beating faster works: running or fast-cycling. 92.15.2.211 (talk) 14:00, 14 October 2010 (UTC)[reply]

Yes, I agree fully, except that walking involves moving the body's centre of mass up and down, and this is the main reason (or one of the reasons) that it uses more energy for a given distance. Some people might interpret "exercise efficiency" as using more energy, not less, so there is vast scope for confusion here. Dbfirs 14:12, 14 October 2010 (UTC)[reply]

(EC) I don't think people are confusing energy and power. People are simply referring to the fact that human energy usage is complicated. Nil Einne (talk) 14:17, 14 October 2010 (UTC)[reply]

As a somewhat OT example of what we're talking about, this study [2] described here [3] found walking 1.6km requires less energy then running that distance. Nil Einne (talk) 14:45, 14 October 2010 (UTC)[reply]

In terms of energy expenditure, cycling lies between walking and running -- where it lies depends on how fast you are riding. On a good road bike, cycling at 12 mph on flat ground, with no wind, is hardly more effort than walking at a brisk pace; cycling at 18 mph is equivalent to running at a modest pace; cycling at 20 mph is equivalent to a pretty fast run. Looie496 (talk) 16:32, 14 October 2010 (UTC)[reply]

True, but cycling requires least energy for a given distance. This article might be of interest for the comparison between walking and running from a mechanical viewpoint. It supports my claim that walking is the least efficient activity, but that is from the physics of the actions. Physiological factors will probably change the balance, and might vary depending on the fitness and running style of the individual. Dbfirs 16:42, 14 October 2010 (UTC)[reply]

Cycling usually allows more distance to be covered and therefore more total energy expenditure, while running would generally use more energy than either walking or cycling. Cycling uphill on a wet muddy gravelly road would use up far more energy than cruising down a steep hill that has a smooth paved surface on a bike for the same distance. ~A H 1^(TCU) 01:42, 15 October 2010 (UTC)[reply]

N-Acetylglucosamine as a protein ligand

These are pretty common ligands for a lot of proteins listed in PDB, especially membrane receptors. Is N-acetylglucosamine used to help induce crystallisation? John Riemann Soong (talk) 11:06, 14 October 2010 (UTC)[reply]

N-acetylglucosamine is a common component of eukaryotic glycosylation, both for N-linked glycosylation an O-linked glycosylation. Cell surface proteins, particularly receptors, tend to be heavily glycosylated. Have you checked if the proteins you thinking of have N-acetylglucosamine as a ligand (non-covalently bound) rather than as a covalent modification? (Most covalent modifications of proteins show up as "ligands" in the PDB files, due to limits of the PDB format.) -- 140.142.20.229 (talk) 23:33, 14 October 2010 (UTC)[reply]

Magnesium flame test

Do magnesium compounds really produce a white flame when it is heated? I heated magnesium oxide, magnesium hydroxide, and magnesium sulfate and none of them made a white flame. They made no flame and didn't even glow red-hot. --Chemicalinterest (talk) 13:53, 14 October 2010 (UTC)[reply]

No. Magnesium metal burns with a bright white flame, but the oxide is just too stable for the compounds to give the same flame colour. Physchim62 (talk) 14:48, 14 October 2010 (UTC)[reply]

Everlasting interest in flame tests

The magnesium oxide you used is also sold as pressed thin sticks to hold them into a flame for flame spectroscopy. This material is a good choice, because it gives no colour when heated and is extremely temperature stable. Platinum is the better alternative, but it is more expensive and for most test magnesia is good enough. --Stone (talk) 18:16, 14 October 2010 (UTC)[reply]

Would carbon work well in a flame test? I recall that it only makes a small flame, similar to copper, and it is unreactive. --Chemicalinterest (talk) 19:11, 14 October 2010 (UTC)[reply]

Pure carbon is to extent already much oxidised; plus it's polyaromatic, making it already quite stable. (It's a powerful testament to the carbocation as a Lewis acid...) When you burn hydrocarbons most of the released energy comes from formation of H2O...the stability of CO2 simply compensates for the loss of C-H bonds. John Riemann Soong (talk) 19:54, 14 October 2010 (UTC)[reply]

Pure carbon oxidized? Isn't it just rows and layers of graphene, carbon? --Chemicalinterest (talk) 00:26, 15 October 2010 (UTC)[reply]

My point of reference was hydrocarbons, sorry! It's C-H bonds that supply most of the energy in a combustion reaction. Carbon dioxide, because of its very electropositive carbon, as you know is a decent Lewis acid -- compare the enthalpies of formation of methane + carbon dioxide (-18 kJ/mol + -393 kJ/mol) versus acetic acid (-490+ kJ/mol). Clearly it's more thermodynamically stable for the carbon dioxide to be reduced to carboxylate by methane (methane gets oxidised by creation of a C-C bond and loss of a C-H bond). To an extent, the second oxygen already has trouble getting electron density out of that carbon, which is why the bond is polar covalent and not say, ionic. John Riemann Soong (talk) 00:50, 15 October 2010 (UTC)[reply]

This brings up an interesting question. How vulnerable would magnesium metal-frame buildings be to fire and corrosion by rain water? What if piping surrounded the building and natural gas filled the pipes? ~A H 1^(TCU) 01:37, 15 October 2010 (UTC)[reply]

This probably isn't a worry as long as magnesium oxide coating survives (though in a fire this might be an issue). John Riemann Soong (talk) 04:51, 15 October 2010 (UTC)[reply]

Chilean miners' dental problems

I understand why skin problems and lung problems have been prevalent during the miners' months underground in 30 degrees (C) temperatures and high humidity, but many reports have mentioned dental problems, too. Two of the miners required immediate dental surgery for abcesses upon rescue. I/we could speculate on everything from poor dental health from the start, through poor nutrition during the ordeal, through poor hygiene (before and during). Can anyone locate hard information on the source being any of the above list or, like the skin and lung problems, a consequence of the underground air and temperature? Thanks from User:Bielle writing from the lobby of a hotel in Ottawa. 207.219.128.198 (talk) 14:57, 14 October 2010 (UTC)[reply]

Professor Francis Hughes (a Professor of Periodontology from Kings College, London) speaking right now on the BBC's PM programme says this is a well-known condition: severe gingivitis known as "trench mouth" (because it was seen in WWI), and ascribes it to poor dental hygene (they couldn't brush for a long time), malnutritition (there was a long time before food could be sent to them), vitamin D deficiency (because of so long without sunlight) and stress (which he said was a major and well-recognised factor for oral abscesses in circumstances like these). -- Finlay McWalter ☻ Talk 16:20, 14 October 2010 (UTC)[reply]

The relevant article appears to be Acute necrotizing ulcerative gingivitis, but I'm eating so I'm too squeamish to read it (lest there be scary photos). -- Finlay McWalter ☻ Talk 16:31, 14 October 2010 (UTC)[reply]

It's safe to look - no photos! AndrewWTaylor (talk) 17:04, 14 October 2010 (UTC)[reply]

If it's so common I wonder why they didn't send down toothbrushes and UV flashlights. --Sean 16:57, 14 October 2010 (UTC)[reply]

I remember reading about toothbrushes when they lowered the first pail lowered down the little borehole, and I'm pretty sure they sent vitamins also. I've seen discussion about whether the form of vitamin D in multivitamins is satisfactory to replace for sunlight or good food, but never came to a conclusion about it. Though I have no evidence at all, I wouldn't rule out that there could be some other photochemical reaction in the skin that is not yet known to science... Oh, and also, melatonin has an effect on the immune system; though the article describes it as positive, sometimes the effects of inflammation are regarded as damaging, and also there tend to be trade-offs in the immune system where stimulating one response may hinder another. Wnt (talk) 17:54, 14 October 2010 (UTC)[reply]

I speculate that there may also be damage arising out of a 70 day coating of abrasive dust, combined with the miners grinding their teeth due to the stress. --Tagishsimon (talk) 18:00, 14 October 2010 (UTC)[reply]

I would suspect stress as they did have access to dental hygiene and their diet was maintained at 2,000 kcal/day. PЄTЄRS J VЄСRUМВА ►TALK 18:12, 14 October 2010 (UTC)[reply]

After the first 17 days that is. Before that, they had like 3 cans of tuna for all 33 of them. Googlemeister (talk)

The first dental problems were beng mentioned in early September, so my speculation is that they were an aggravation of pre-existing conditions rather than something completely new. So some of the miners had less-than-perfect teeth to start with, and these problems got worse (for the reasons described above) to the point of being really troublesome. Physchim62 (talk) 01:46, 15 October 2010 (UTC)[reply]

While I don't wish to shut down any other explanations or references, I do want to thank those who have already replied. I have learned a lot. 207.219.128.198 (talk) 19:18, 15 October 2010 (UTC)[reply]

Semi-terrestrial octopuses/octopi/octopodes...

Just been reading the Pacific Northwest tree octopus (which is a hoax) article and it brought to mind video footage I've seen of octopi (or whatever) actually moving reasonable distances across land, say to escape from an isolated pool in which they have become trapped when the water receded. Which species in particular have this ability? --Kurt Shaped Box (talk) 18:14, 14 October 2010 (UTC)[reply]

This thread on the topic mentions several species. --Sean 18:25, 14 October 2010 (UTC)[reply]

Also look at tidal pool. ~A H 1^(TCU) 01:32, 15 October 2010 (UTC)[reply]

How does this levitation magnet system's feedback system likely work?

This one: http://www.crealev.com/wp-content/uploads/A4_CLM_2.pdf Thanks 20.137.18.50 (talk) 18:43, 14 October 2010 (UTC)[reply]

The literature notes optical sensors; I guess these feed positional information to a controller which varies the strength (or duration) of each one of a set of electromagnets in the base unit. --Tagishsimon (talk) 21:22, 14 October 2010 (UTC)[reply]

A single electromagnet could suffice if its core is shaped to give a suitable shape of external field. The reference does not specify the stability of the levitation height so it is worth noting that one could obtain a similar levitation without feedback, or even without electric power if permanent magnets are used. The levitation height would then decrease with load. Cuddlyable3 (talk) 09:57, 15 October 2010 (UTC)[reply]

No it can not be done with permanent magnets only see Earnshaw's theorem and Magnetic levitation. --Gr8xoz (talk) 11:16, 15 October 2010 (UTC)[reply]

There are several exceptions to Earnshaw's theorem that allow magnetic levitation, such as when the levitated material is diamagnetic. (The system in the reference may not be rotationally stable.) Cuddlyable3 (talk) 13:12, 15 October 2010 (UTC)[reply]

If diamagnetic materials are used then the levitation is not done with permanent magnets only. Diamagnetic effects are very small except for cryogenic superconducting materials so it is very unlikely that it would be able to lift 10 kg with permanent magnets and diamagnetic effects. Of curse it need to be 'rotationally' stable, in order to levitate it can not tip over so it need to be stable around two axes. The rotation around the vertical axis need to either have zero moment at all angles (It is on the stability limit) or it need to have both positive and negative moments over a turn in order to conserve the energy, and in that case there must be an angle of rotation that gives rotational stability. --Gr8xoz (talk) 14:17, 15 October 2010 (UTC)[reply]

Horse Chestnuts, leaf miner and bleeding canker

I have just seen a BBC nature program implying (contrary to what I have previously read) that the leaf miner affecting most British horse chestnuts is responsible for materially weakening the tree leading to infection with the fatal "bleeding canker". Is there any more authoritative source on this link? It matters because I have a number of red-flowered horse chestnuts which appear immune to the leaf miner and I need to guess whether that species is vulnerable to the bleeding canker or not; to decide whether to plant them as a replacement for the white flowered horse chestnuts which are dying. None of the reds have any sign of canker or a single leaf mined. --BozMo talk 21:28, 14 October 2010 (UTC)[reply]

We have an article on the Horse-chestnut leaf miner. You may be able to find some information there. --Jayron32 22:23, 14 October 2010 (UTC)[reply]

Personal research, but I am not convinced these trees are dying. The problem is in the leaves and infestation does not start until the leaves are mature. The browning occurs from midsummer and worsens into autumn thus giving the tree a significant amount of time to 'thrive' before the leaves are affected. The leaves then drop off and the horse chestnut miner with them. The trees I have examined do not not have any sign of 'canker' and the size of the 'conkers' do not indicate a problem in the plants ability to produce seed. This problem does not bear any comparison with the Dutch Elm disease of decades ago. Essex County council are not too concerned at present[4]My opinion is the BBC Autumnwatch were being a bit over the top about the fate (or success) of the horse chestnut trees. Richard Avery (talk) 07:28, 15 October 2010 (UTC)[reply]

There was a discussion on this a few weeks ago. Alansplodge (talk) 08:43, 15 October 2010 (UTC)[reply]

Odds of discovery of Pioneer plaque by an ET

Of the two 'messages to ET' sent out, the Pioneer plaque and Voyager Golden Record (were there others?) what are the odds that either one of these would be discovered by an extra-terrestrial intelligence? And what is the most likely way it would be discovered? ie. is it most likely that the probe would by chance crash into a planet? or is it more likely that it would it be found floating in space by a space ship? or will it just continue 'voyaging' for millions of years never hitting another another cosmic object? -- œ ^™ 21:39, 14 October 2010 (UTC)[reply]

Voyagers are now in their "interstellar" mission phase. "In about 40,000 years, Voyager 1 will drift within 1.6 light years (9.3 trillion miles) of AC+79 3888, a star in the constellation of Camelopardalis. In some 296,000 years, Voyager 2 will pass 4.3 light years (25 trillion miles) from Sirius, the brightest star in the sky." You can read about SETI and the Drake Equation to inform your speculations about the probabilities of encountering intelligent life in the universe: "unlikely" sounds like a reasonable prediction for a "face-to-face" or orbital-rendezvous encounter between Voyager and some life-form. For perspective, imagine if a small extra-terrestrial craft passed 1.6 light-years from Earth - its RADAR and optical cross-section would be essentially zero; its presence would be transient; and if it happened to pass in, and then out, of range even just 300 years (or heck, even one second) too early, intelligent species who were actively looking for it might not be operating the technology for a deep sky survey to find it. So, it's most probable that Voyagers and Pioneers will drift indefinitely, probably for millions or billions of years, before their materials degrade beyond recognition as a "technological object". Nimur (talk) 22:12, 14 October 2010 (UTC)[reply]

Or the Americans go collect them in a couple of hundred years and put them on public display in the National Air and Space Museum. --Kurt Shaped Box (talk) 22:26, 14 October 2010 (UTC)[reply]

Have you estimated the date of launching a US salvage mission and the accelerations necessary to achieve that collection. Would this be a manned vehicle? Cuddlyable3 (talk) 09:43, 15 October 2010 (UTC)[reply]

(after edit conflict.)

Tough to say. For all we know we might recover Voyager and Pioneer some day and put them in a museum. If space travel that is even remotely cheap is invented, all those old probes will probably be tracked down by treasure hunters. While still a bit far fetched, to me that seems far more likely than an ET recovering them. Sure they're moving away from us, but at least we know where to look for them.

Active SETI projects have beamed messages to (potential) aliens. They probably have a higher chance of being noticed, but still quite low. APL (talk) 22:35, 14 October 2010 (UTC)[reply]

Not factored into the above estimates are the probability that some kind of space faring alien civilization has advanced means of differentiating between space junk over long distances. "Say there, Rhsdfndfpop, the super-duper-long-range-asteroid-tracker (SUDULORAST) has found something that looks extraterrestrial in origin out there near Sirius. Send our probe out there at warp 400 and pick it up." Could happen. Impossible to really say how likely it is — we simply lack sufficient data to make a very good "likely" judgment. I think we could say that an alien with current Earth knowledge, or even Earth knowledge extrapolated for 100 years or so, has essentially zero chance of finding it. But if we allow for not-totally-prohibited-by-physics possibilities, it might be more feasible, if there is some kind of "search and find" possibility. I say this though as someone who reads the Drake equation very pessimistically when it comes to the possibility of human-ET contact. --Mr.98 (talk) 22:41, 14 October 2010 (UTC)[reply]

Unless they have some new kind of physics for detection, we can actually say that can't happen. The diffraction limit pretty much rules out any possibility of this, unless they have planet sized detectors, but worse is that within a hundred years or so the probes will not be radiating anything, and they will be basically invisible. Ariel. (talk) 23:59, 14 October 2010 (UTC)[reply]

There are still the unknown possibilities that they have cleverly thought of a way to use physics as we understand it to build a SUDULORAST, or that physics as we understand it is merely approximation of reality and the difference allows the possibility of a SUDULORAST.

The first is pretty unlikely, but we can't rule out the possibility that alien ways of thinking and alien ways of doing things make certain solutions far more likely to be discovered than our human way of thinking does.

The second is even more complicated. Physics as we know it almost certainly is an approximation for something else, but what? There's simply no way to guess without actually discovering it. Two hundred years ago no one could have sat around pondering the probability that one day special relativity would be discovered by an alien race. APL (talk) 00:51, 15 October 2010 (UTC)[reply]

Who is ruling out planet-sized detectors? These guys are aliens! Maybe they're into that kind of thing. Maybe they got clobbered once with an asteroid a long time ago and decided that they'd take the nearest rock planet and coat it with nanobots who could magically terraform the whole thing using only solar power and naturally occurring methane and self-organize into a planet-sized detector. --Mr.98 (talk) 01:28, 15 October 2010 (UTC)[reply]

Nobody is ruling them out. In fact reputable peer-reviewed journals have published papers suggesting that we should keep an active watch for astrophysical evidence of planetary-sized and solar-system-sized construction activity by advanced civilizations: Search for Artificial Stellar Sources of Infrared Radiation was published in the journal Science, and Communications from Superior Galactic Communities was published in Nature. Nimur (talk) 02:44, 15 October 2010 (UTC)[reply]

I guess the least unlikely scenario is that ET monitors Earth and discovers from us that we sent them. Maybe they are already in ET's Earth museum or on the way and we are only receiving fake signals from ET's replacement. PrimeHunter (talk) 05:01, 15 October 2010 (UTC)[reply]

Based on NASA tracking data to the point the signal is finally lost, the trajectory through interstellar space might be well defined enough that alien civilizations monitoring our computer networks now or in the future might send out a craft to retrieve Earth's first two interstellar probes and return them to us after they reveal their existence. Or maybe they monitored the launch and tracked it with their superior systems. When a baby drops his rattle, an adult may similarly say "Uh oh, dwopped it!" and hand it back to the baby. Edison (talk) 05:23, 15 October 2010 (UTC)[reply]

The Arecibo message was a putative message to aliens transmitted in 1974. Nobody is holding their breath expecting a reply soon. Cuddlyable3 (talk) 09:47, 15 October 2010 (UTC)[reply]

First, we assess the probability that there are extraterrestrial civilizations in the vicinity. If we believe that our civilization has a good chance to survive and grow, so should we assume that a significant proportion of the extraterrestrial civilizations do too. With the current increase in our energy consumption, 3% per year, we will use much of the sun's total output in 1000 years. (Not just the part that hits the earth but the whole power of 3.846 × 10 ^ 26 W) The long-wave infrared radiation which we will emit will make us easy to detect.

In order to continue to grow as a civilization, we must either increase the solar power or colonize other stars. It is reasonable to assume that it takes less than 100 milijoner years to colonize most of the galaxy. (Travel by about 0.1% of the speed of light) Long before this the civilization should be very easy to detect. Civilizations like ours has been able to evolve in at least the past 2 billion years. Given that we have still not seen any signs of extraterrestrial civilizations in either the Milky Way or in other galaxies this suggests that they are very rare. I would guess on between one civilization in the observable universe to a few small isolated civilizations in each galaxy.

Pioneer plaque and Voyager Golden Record will with very high probability continue their journey in millions of years unless we retrieve them before then, the question is what will happen to them for billions of years. It seems unlikely that they will be detected over interstellar or intergalactic distances, although it is difficult to predict the detection ability of extraterrestrial civilizations. They will only go near (<100 AU, 0.00158 light yers), a few stars (less than one star per billion years.). The probability that these have civilizations is very low. A civilization must be much more advanced than ours to detect that they are artificial at interplanetary distances. The probability that they fall down on an inhabited planet must be close to zero. The greatest likelihood that they will be found by extraterrestrial civilizations is probably that they get information from us about where to look by radio or otherwise.

A civilization that colonizes a large part of the galaxy would have a reasonable chance of finding them but because we do not see such civilisations in other galaxies, it seems unlikely that any other civilizations than our will colonize the Milky Way. It is very unlikley that the probes will be found by planetary civilizations like ours whithout information of where to look. I would put odsen higher than one to 100 billion. With extraterrestrial civilizations, I mean civilizations that arose independently of life on Earth. Relevant articles are Fermi paradox, the Drake equation and Dyson swarm. --Gr8xoz (talk) 13:53, 15 October 2010 (UTC)[reply]

Not to be rude, but just because our civilization is currently expanding our energy usage by 3% a year does not make it statistically significant. You have a sample size of 1. Googlemeister (talk) 18:57, 15 October 2010 (UTC)[reply]

For the relevant discussion of this energy-growth, see our article on the Kardashev scale. Any discussion of SETI is limited by our "sample size", but we also know some things about basic mechanics and energy. It has been brought up in many SETI contexts that our definitions for "life" and "intelligence" are just subsets of generalized energy transfer/transformation processes. If our Voyager probe were to encounter a dense hydrogen cloud, some billion years in the future, in an unlikely region of space, could we safely say it has encountered a "natural phenomenon" as opposed to some intelligent natural process? Exactly how sophisticated does a dense accumulation of atoms need to be before we consider some portion of it "intelligent"? It is fair to say that in our conventional wisdom, some portion of those atoms need to be undergoing stellar fusion, and some other portion of those atoms need to be harnessing the energy that stellar fusion converted into electromagnetic radiation, and repackaging it as complex chemicals; and some other portion of those atoms need to be eating those chemicals so that they can re-invent stellar fusion without obliterating themselves. (We call the last group of atoms intelligent until they obliterate themselves). If those criteria are sufficient for intelligent life, any sufficiently dense blob of hydrogen might qualify as intelligent life, depending on conditions. Nimur (talk) 19:37, 15 October 2010 (UTC)[reply]

Thank you all for the excellent answers, I am satisfied. I'm left fascinated with the thought of these spacecraft, if not recovered by us sometime in the future, continuing on travelling indefinately through space, not encountering any other object because of the vast distances between the stars. Yet knowing this scientists still held out enough hope to place these ET messages on them. I can only hope this optimism grows with time. -- œ ^™ 02:49, 16 October 2010 (UTC)[reply]