Wikipedia:Reference desk/Archives/Science/2011 November 26

= November 26 =

Newtonian gravity and relativistic mass
Does Newton's law of gravitation abide by the notion of relativistic mass? In other words, if for an object moving at velocity $$v$$ and rest mass $$m_0$$, since its relativistic mass is given by $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$, would the gravitational acceleration it would impose be of magnitude $$a_g=\frac{Gm_0}{r^2\sqrt{1-\frac{v^2}{c^2}}}$$ for some spatial separation $$r$$? I know numerous suggest against the interpretation of relativistic mass. Is the possible false nature of this statement one of those reasons? Does this have a known answer? Thanks! — Trevor K. — 00:29, 26 November 2011 (UTC)  — Preceding unsigned comment added by Yakeyglee (talk • contribs)
 * This is a good example of how the concept of "relativistic mass" can be very misleading. Newton's gravitational law is valid for a source mass that is at rest, so it depends on the rest mass. Computing the gravitational field of a mass at high speed requires general relativity, but for the moment let's just assume that we can get it from a Lorentz transform of the Newtonian field. I do not know exactly what the result of that exercise is but the field will definitely become anisotropic due to relativistic and retardation effects, hence it will look different along the direction of motion and perpendicular to it, and presumably also different ahead and behind the mass (one can look at electrodynamics for inspiration). Whatever the precise shape may be, it certainly cannot be described by changing a single scalar quantity, namely rest mass to relativistic mass. It can be described by using the appropriate coordinate transformation, taking into account magnitude and direction of the velocity vector. --Wrongfilter (talk) 12:14, 26 November 2011 (UTC)


 * Luckily Einstein did conveniently provide a gravitational theory that is consistent with his special theory of relativity. — Preceding unsigned comment added by 129.67.38.22 (talk) 11:44, 28 November 2011 (UTC)
 * In case the above responses leave any doubt, note that the gravitational effect of a moving body does increase with velocity, it's just that you can't get the correct result by plugging the so-called "relativistic mass" into the classical equation. You have to use General Relativity.--Srleffler (talk) 17:38, 28 November 2011 (UTC)

Calculating latitude via Earth's celestial rotation speed
Hi. Recently I calculated that Jupiter's movement in an equatorial telescope across the egde of the FOV was approximately 4.76 seconds, i.e. it took that length of time between one edge of the planet leaving the FOV to the entire planet exiting it. The Earth's rotation is of course magnified for celestial objects, and assuming that Jupiter's disk is exactly 45 arcseconds in angular diameter, I calculated Earth's rotational speed for Jupiter to be approximately 9.45 arcsec/s, or 9.45 deg/h. Assuming that one Earth day relative to the celestial sphere is 24.0 hours (and not 23 hours 56 minutes), Jupiter is moving across the sky at 226.9 deg per day. Now, this is likely accurate, because one would expect an object on the celestial equator to move 360 deg. per day, and with all degrees at the same angular size objects away from the equator would move more slowly. By contrast, a bright star exactly on the celestial pole (Polaris is close, but not quite on the pole) would move 0.0 deg per day. Dividing Jupiter's angular movement number by π gives 72.2 deg, or 1.26 rad. For an equatorial object, this number would be 2.00 rad. We now have the ratios. This is not homework. My question is: how can we calculate the exact angle in degrees an object such as Jupiter is above or below the celestial equator (the declination) based on this data? I know that superimposed on a flat circle, the ratios can be represented as 2.00 for the diameter and 1.26 for the length of a chord parallel to the diameter line&mdash;how would we calculate the angle between the diameter line and a line connecting the origin of the circle to the point where the chord meets the edge of the circle? Furthermore, how can one use the declination and the observer's latitude to calculate the time that an object would spend crossing the sky, from an altitude of 0 to an altitude of 0? Also interested in using that information to calculate the maximum altitude in degrees an object would appear in the sky, where the maximum occurs when the object transits the line going from celestial north to celestial south, and the applications to solar elevation angle, determining the hour of day relative to solar noon based on the available data. Trigonometry is likely to be used, with or without calculus. Thanks. ~ AH1 (discuss!) 00:54, 26 November 2011 (UTC)


 * You're 90% there! The angle is simply $$\arccos(1.26/2)$$ = 51°. Jupiter's declination is currently about 10°, so unfortunately your estimate is not very good. The procedure assumes a fixed mount, not an equatorial mount. Are you sure that the motor was off? For your other questions, your zenith vector as a function of time is $$(\cos(t) \cos(lat), \sin(t) \cos(lat), \sin(lat))$$, while the object can be assumed fixed at $$(\cos(dec), 0, \sin(dec))$$. Read the first 2 sections of dot product and see how far you can get. Note that both are unit vectors. 98.248.42.252 (talk) 06:36, 26 November 2011 (UTC)

Observing light indirectly
is it possible to "observe" light indirectly ? it is possible to tell if light is crossing a vacuum without somehow interfereing with its passage in order to determine if it is there or not ? — Preceding unsigned comment added by 157.130.137.234 (talk) 01:58, 26 November 2011 (UTC)
 * I added the missing section title. -- ToE 03:25, 26 November 2011 (UTC)
 * If there is enough of it it will distort the fabric of space causing gravity. Appearing as something like dark energy.  If the energy is high enough eg UHEGR the photons may interact with cosmic background radiation forming other photons or particles. Graeme Bartlett (talk) 05:29, 26 November 2011 (UTC)
 * Gamma ray is a section that talks about this. But this is photons interacting with photons. An Extra-galactic pair halo is theoretical formed around high energy gamma ray sources with energy over 10TeV. This halo reveals the amount of low energy photons in the region. Graeme Bartlett (talk) 12:27, 26 November 2011 (UTC)


 * so, in other words, light has gravity, because light has mass so it is possible to determine the presence of light crossing a vacumm by observing a distortion through/ or around it ?


 * According to quantum mechanics, the answer is a solid NO. It is fundamentally impossible to measure/observe light without altering it. To the extent that this disagrees with the general relativistic response above, note that general relativity and quantum mechanics are fundamentally incompatible.--Srleffler (talk) 17:41, 28 November 2011 (UTC)
 * Wait. If you're observing the light because it's so dense it pulls you by gravity, then you are modifying the light because you are also pulling it.  Quantum mechanics doesn't say that's impossible.  Similarly, if you are observing light because the photons are reacting with each other and generate particles which then fly to you, that's also not against quantum mechanic.  The “general relativity and quantum mechanics are fundamentally incompatible” part is completely irrelevant here.  &#x2013; b_jonas 22:00, 30 November 2011 (UTC)

Non traditional mosquito repelent
Living in Mississippi, near the water, you can imagine we have problems with mosquitoes. I have seen some of my friends hanging plastic (ziplock) bags filled with water to repel the mosquitoes. Honestly, I thought this was bullshit, but I tried it on my patio, and I'll be damned if it doesn't work fantastically. What is behind this idea of suspended water repelling mosquitoes? Quinn &#10025; STARRY NIGHT 03:39, 26 November 2011 (UTC)


 * According to Uncle Cecil's Staff Reporters, this is supposed to work for flies and not mosquitoes. Absent any other evidence, I suspect that the result is some combination of confirmation bias and the placebo effect.  I.e. it doesn't really work, but you believe it to work, so you have convinced yourself that it does.  I'm still looking, but haven't found anything I would call reliable to explain the supposed phenomenon.  -- Jayron  32  03:53, 26 November 2011 (UTC)


 * Here are the results of a field trial (for flies) by Mike Stringham, professor of entomology at North Carolina State University.  Sean.hoyland  - talk 03:59, 26 November 2011 (UTC)


 * Interestingly enough, his results indicate that plastic bags of water attract flies; meaning that not only does the supposed remedy NOT work, it actually does the exact opposite of what it is supposed to do. Good stuff.  -- Jayron  32  04:06, 26 November 2011 (UTC)


 * The human vs mosquito/fly wars are an endless source of entertainment. Just the other night, having patiently collected a great deal of pomelo peel for many weeks, we were out in the garden testing whether burning the peel reduced the presence of any of the wide variety of mosquito species we have available as test subjects. Apparently not, but we did establish that the burning pomelo peel is very pretty and smells nice. I'm curious about how many small black puppies and/or kittens per cubic metre are required to reduce the mosquito bite frequency for a cohabiting human to zero. There were some ethical problems with the suspended black puppy and kitten field trial.  Sean.hoyland  - talk 05:46, 26 November 2011 (UTC)


 * In southern Spain I have seen people putting 2 litre plastic drinks bottles full of water at 3 or 4 metre intervals around their house in the firm belief that it will deter ants from entering the house. I have been unable to find any scientific work into this most interesting idea! @Jayron above plastic bags of water may attract flies and this would work as a distraction strategy, like a hanging jar of jammy water attracts wasps (yellowjackets) away from a meal table. Maybe. Richard Avery (talk) 08:01, 26 November 2011 (UTC)


 * Mosquitoes and other flying insects in colder climates are also seasonal. That may have something to do with the apparent ceasing of the swarms. That said, there are claims that it may also be caused by sunlight refracting off the water bags, giving the illusion of movement to insects like flies which have highly acute vision.--  Obsidi ♠ n   Soul   09:34, 26 November 2011 (UTC)


 * Perhaps warm water might attract the mosquitoes, and, thinking they found their prey, they get a snoot full of water instead. StuRat (talk) 01:27, 27 November 2011 (UTC)


 * Wouldn't you have to have a source of warm and humid carbon dioxide if you really wanted a "distraction"?
 * --DaHorsesMouth (talk) 05:06, 27 November 2011 (UTC)


 * I would guess that it is humid around a plastic bag full of warm water, since some water vapor passes through. If you want to add carbon dioxide, I believe there is a premium mosquito catcher that burns fuels to generate that, as well as water vapor and heat, and catches thousands of skeeters.  I doubt it plastic bags full of warm water work anywhere near as well, but enough of them might still have some effect. StuRat (talk) 16:55, 27 November 2011 (UTC)

Freezing pop
Let's say you buy some pop at the store and then put it out in your garage. It stays out there for some time and the temp in the garage gets below freezing. You then bring a bottle into your house and open it up. When you do, ice crystals start to form in the bottle after you open it. What's going on here? Why does the pop start to crystallize/freeze? Was the pressure inside the bottle the same as it was when initially bottled? This isn't a homework question. I'm not in a physics class or anything. Another person and I are just having a disagreement about what causes this and what the pressure in the bottle was and now is (relative to what it was immediately before opening). Thanks, Dismas |(talk) 04:44, 26 November 2011 (UTC)
 * See Adiabatic_process. The opening of the bottle causes a rapid drop in pressure.  As the pressure decreases, the temperature of the liquid drops, not due to any release of heat, but due to the work the expanding gas does on its surroundings (i.e. it is an "exergonic" process rather than an "exothermic" process).  Since the work done on the surroundings has to come from somewhere, it comes from the temperature of the liquid.  If such work lowers the temperature below the freezing point, the liquid will freeze.  This is pretty much exactly how a refrigerator works.  The pressure in the bottle will drop slightly due to changes in environmental temperature, but the difference between room temperature and the temperature in your garage isn't going to lower the pressure a whole lot, on the order of maybe 10% or so.  The greater effect on pressure is merely opening the canister.  -- Jayron  32  05:05, 26 November 2011 (UTC)
 * @Dismas: I see you are a transplant. Someone who is a Vermont native would never talk about freezing pop. Magog the Ogre (talk) 06:11, 26 November 2011 (UTC)
 * I had to think twice to figure out that Dismas wasn't freezing Grandpa. Is pop all that common as a name for the drink in the USA these days? HiLo48 (talk) 06:17, 26 November 2011 (UTC)
 * I believe in Vermont they freeze tonic. At least, when I was growing up in New Hampshire, that's what the bubbly sugary stuff in the bottles and cans was called.  -- Jayron  32  06:19, 26 November 2011 (UTC)
 * I believe "pop" is not uncommon in the Buffalo and general upstate/western New York region. It was quite common when I lived there anyway. It's not common on the west coast. Pfly (talk) 07:12, 26 November 2011 (UTC)
 * It's very common in the American midwest. Short for "soda pop". ←Baseball Bugs What's up, Doc? carrots→ 13:48, 26 November 2011 (UTC)

Yes, I meant soda pop. And yes, I am from the Midwest. Thanks for the answer and links, Jayron! Dismas |(talk) 16:47, 26 November 2011 (UTC)

There's a soda–pop line somewhere in the middle of the country. Here in California it's definitely soda or soda pop, never pop alone. I think Wikipedia may actually have a map showing the regional variations (in third place is probably Coke for all fizzy soft drinks whether they come from the Coca-Cola company or not &mdash; I think that's mostly in the South,) --Trovatore (talk) 19:59, 26 November 2011 (UTC)


 * Ask and ye shall receive, though it does not include some terms, like the aformentioned "tonic", though that term may be becoming archaic; it seems to be being used less and less in New England. -- Jayron  32  20:05, 26 November 2011 (UTC)
 * That's the one I was thinking of, thanks. --Trovatore (talk) 20:15, 26 November 2011 (UTC)
 * There's also this map--and lots of other curious dialect-related maps at the same site, http://dialect.redlog.net/maps.html (like, I had never heard someone refer to a bubbler until I married a Bostonian). Pfly (talk) 07:06, 27 November 2011 (UTC)
 * Weirdly enough, that use of bubbler exists in Australia too, but it's not all that common. HiLo48 (talk) 07:12, 27 November 2011 (UTC)


 * According to supercooling this phenomenon is reliable enough that Coke actually developed a specialty vending machine that would deliver cans that would turn to slush upon opening. I assume that the bubbles forming within the liquid when the pressure is relieved are what nucleate the ice formation. Wnt (talk) 02:33, 27 November 2011 (UTC)
 * Wow, that would actually be a three-phase system -- gaseous CO2 bubbles dispersed inside solid ice crystals, which are themselves suspended in liquid Coca-Cola. :-) 67.169.177.176 (talk) 08:25, 27 November 2011 (UTC)
 * I don't think it works like that. Nucleation involves somehow nudging a tiny region into the right configuration - the classic chem lab example is scratching the inside of a beaker with a glass rod to start the crystallization of something dissolved in solution.  Only one or a few nucleation events occur, then the stray bits of crystal spread around in the liquid and get the rest to come out.  I think sometimes you'll see multiple crystals started along the scratch, but it's some countable number.  So I think in the cola example you'd have just one or a few bubbles starting nuclei, and they probably go their separate ways before the freeze spreads to the rest of the liquid.  There would be three phases as you say, but the bubbles and ice crystals wouldn't have any special relationship. Wnt (talk) 18:56, 30 November 2011 (UTC)


 * Couldn't find the on-wiki map, but here is one showing the American distribution of pop-vs-soda-vs-coke. Freezing pop was how the popsicle was invented. ~ AH1 (discuss!) 02:42, 1 December 2011 (UTC)

Dark Matter questions
I watched a program tonight on dark matter, and it's left me with a few questions: Magog the Ogre (talk) 05:47, 26 November 2011 (UTC)
 * Apparently there are detectors deep underground waiting for the collision of a WIMP with an atom (see Dark matter). But if a WIMP doesn't interact via any force but gravity, then why would it interact with another particle in any observable way at all? Is the experiment assuming that it gets so close to an infinitesimally small particle that it moves the particle via gravity? Wouldn't the particle zip by so fast that even if it came within Planck length of non-WIMP particle, that it would fail to move that particle? (Pardon my quantum physics ignorance; I'm 100% self-taught).
 * If dark matter exists throughout galaxies, and it is composed of WIMPs, doesn't this mean that dark matter should collect at the center of stars? If matter collects in a star due to gravity, so should dark matter. Then, when the stars go supernova, it would leave a gigantic clump of dark matter just sitting there at the center, making any of the star's remnant far heavier than it should rightly be.
 * Why isn't the effect of dark matter visible in the solar system? How do we know that our readings of the gravitational constant have not been affected by dark matter at the center of heavy bodies?


 * The first paragraph of our WIMP article says that "these particles interact through the weak force and gravity, and possibly through other interactions no stronger than the weak force", so it's not true that a WIMP doesn't interact via any force but gravity. Matter collects in a star due to gravity and friction. When dark matter enters a body, it exits at the same speed so it doesn't collect. That said, there might be a tiny accumulation due to n-body effects. 98.248.42.252 (talk) 07:27, 26 November 2011 (UTC)


 * Dark matter does collect at the center of stars due to the interaction with the nuclei in the star. The dark matter in the Sun will then self-annihilate into high energy neutrinos, these can in principle be detected. This is the basis for the so-called indirect dark matter searches. Dark matter will also collect at the center of the Earth, but it turns out that the neutrino signal from this will be less strong than from the Sun. Count Iblis (talk) 15:27, 26 November 2011 (UTC)

re 98.248: right. As soon as I walked away it occurred to me the silliness of it. What I was referring to was particles that react only through gravity. Because these phantom particles would have mass, they could be slowed down by the presence of a gravitational object. Magog the Ogre (talk) 19:37, 26 November 2011 (UTC)

re: Count Iblis: what proof is there that the dark matter would degrade into neutrinos? Magog the Ogre (talk) 19:38, 26 November 2011 (UTC)


 * This is if DM consist of neutralinos . Count Iblis (talk) 19:56, 28 November 2011 (UTC)


 * The Sudbury Neutrino Observatory is one such detector. ~ AH1 (discuss!) 02:34, 1 December 2011 (UTC)

science
why does water not enter an empty glass when it is carefully inverted in a owl full of water ? — Preceding unsigned comment added by 182.179.73.187 (talk) 06:30, 26 November 2011 (UTC)
 * Because you have to get the air out of the glass first. -- Jayron  32  06:31, 26 November 2011 (UTC)
 * An interesting extension is that if you have a very tall glass, even if you could fill it completely with liquid water and then invert it into a bowl, there might still be a visible gap. That's the principle of a barometer. DMacks (talk) 09:25, 26 November 2011 (UTC)
 * Could you elaborate on that? Where does the air in your gap come from if the glass is completely full of water? A barometer needs an air gap to start with, which then changes size as the pressure changes. --Tango (talk) 14:13, 26 November 2011 (UTC)
 * (EC) No, it doesn't. The gap at the top of a classic fluid barometer (such as DMacks describes) is not an air gap, it's a vacuum (or at least it starts as one - it may contain some vapour of whatever the working fluid is). {The poster formerly known as 87.81.230.195} 90.193.78.52 (talk) 14:55, 26 November 2011 (UTC)


 * There are a number of ways to construct a barometer. Aneroid barometers contain one or more sealed metal capsules full of air that expand and contract with changes in the pressure difference between their interior and exterior.  Similarly, a weather glass contains a chamber filled with air, sealed with a plug of liquid water in a tube; the water level rises or falls with changes in the relative pressure.
 * The other major way to build a barometer is to require the ambient pressure to support a column of liquid against its own weight. In the classic mercury barometer, a tube filled with liquid mercury and sealed only at its top end is immersed in a pool of mercury.  The mercury in the column will fall under the influence of gravity (forming a near-vacuum at the top of the tube) until the pressure of the liquid mercury at the open bottom of the tube matches the air pressure on the surface of the surrounding liquid.  Under standard conditions, the ambient air will support a column of mercury 760 millimeters high (it is from this physical measurement that we derive the millimeter of mercury – mmHg, or torr – as a unit of pressure).  Water columns, being much less dense than mercury, can be supported to much greater heights by atmospheric pressure: about 10 meters, or 30 feet.  (By the same principle, this height also represents the maximum vertical distance that you can lift water by suction; even with a perfect vacuum pump at the top of the pipe, you wouldn't be able to suck water uphill more than 10 meters.  This also has practical implications when attempting to siphon liquids over a tall barrier or height of land.) TenOfAllTrades(talk) 14:52, 26 November 2011 (UTC)

--TammyMoet (talk) 09:53, 26 November 2011 (UTC)
 * Oh, give it a try. It would be a hoot. ←Baseball Bugs What's up, Doc? carrots→ 13:46, 26 November 2011 (UTC)

Anti-gravity properties of jam
Purchased a jar of jam and some was removed for consumption purposes. The jar was then put into a packsack and taken 2.5 km before being placed into a fridge. Upon opening the fridge the jam was found with a 4 cm gap at the bottom. See picture on the right. What would cause that? CambridgeBayWeather (talk) 08:51, 26 November 2011 (UTC)


 * The obvious (it was turned upside down) and the arcane (patent it quick! We'll have flying cars running on jam by 2012).--  Obsidi ♠ n   Soul   09:16, 26 November 2011 (UTC)


 * Sorry forgot to mention that's what I thought at first but this was taken two days after the jam had been right side up. CambridgeBayWeather (talk) 10:06, 26 November 2011 (UTC)


 * Could it be because it's behaving as a Bingham fluid i.e. while it was being transported the vibrations or whatever provided enough shear stress for it to flow but once that stopped it switched to rigid body behavior ?  Sean.hoyland  - talk 12:55, 26 November 2011 (UTC)
 * Also it's viscosity would have changed from when it was in the warm packsack to the cold fridge. Maybe just enough to allow the formation of this relatively stable state.  If so, you would expect it to drop eventually. 129.234.53.239 (talk) 13:57, 26 November 2011 (UTC)
 * Which was How Tom Beat Captain Najork and his Hired Sportsmen, although in that case the power came from jam's reaction with a new material called anti-sticky. Card Zero  (talk) 19:26, 26 November 2011 (UTC)
 * I suspect it was turned upside down gradually (or, perhaps, just in two stages - on to its side and then on to its top). That allowed the air above the jam to get to the other side of it. You've turned it the right way up in one go, which means the air is now trapped underneath. If you get a knife and make a small hole through the jam, you'll find it suddenly drops down as soon as the air-tight seal the jam is creating is breached. Alternatively, leave it on its side for a few hours until the jam flows down to the side of the jar and then turn it the right way up. --Tango (talk) 14:17, 26 November 2011 (UTC)

Thanks all. We were able to get most of the jam out before it dropped. CambridgeBayWeather (talk) 16:47, 27 November 2011 (UTC)


 * So what happened? What did you end up doing? Did the make-an-air-hole technique work? If so, how quickly? This is important for science!!! SamuelRiv (talk) 20:36, 27 November 2011 (UTC)


 * We just used it up as per normal. Carrying it back and forth in the packsack made no difference to it. After a couple of days we went through the bottom and the jam dropped. CambridgeBayWeather (talk) 05:16, 28 November 2011 (UTC)

If you left it long enough for the jam to fall on its own, the process by which the jam breaks through the air holding it up is the Rayleigh–Taylor instability.--Srleffler (talk) 17:52, 28 November 2011 (UTC)

Deinosuchus skull reconstruction

 * Transplanted from Help desk

Hi,

There is a problem with this part. Once the skull is to broad, and than it is not enough broad. Someone please explain.

The American Museum of Natural History incorporated the skull and jaw fragments into a plaster restoration, modeled after the present-day Cuban crocodile.[7] Colbert and Bird stated that this was a "conservative" reconstruction, since an even greater length could have been obtained if a long-skulled modern species such as the saltwater crocodile had been used as the template.[7] Because it was not then known that Deinosuchus had a broad snout, Colbert and Bird miscalculated the proportions of the skull, and the reconstruction greatly exaggerated its overall width and length.[11] Despite its inaccuracies, the reconstructed skull became the best-known specimen of Deinosuchus, and brought public attention to this giant crocodilian for the first time.[11] — Preceding unsigned comment added by 92.84.199.195 (talk) 12:07, 26 November 2011 (UTC)


 * The answer is simple: as with a lot of fossils, the original remains of Deinosuchus were very fragmentary. Paleontologists can only make guesses as to what it actually looked like until more complete specimens can be recovered. There are numerous examples of hits-and-misses like this, such as Iguanodon originally being depicted with their thumbs on their noses (literally), Stegosaurus with their bony flaps laid flat against their backs, and Velociraptor depicted with no feathers.


 * As more specimens are recovered, the reconstruction becomes more accurate. Even if also fragmentary, additional specimens (paratypes) can contribute to reconstructing the whole animal's appearance, provided that they are identified correctly (by characteristics shared by specimens from two different individuals for example).


 * In Deinosuchus, the size of the fragments led paleontologists to overestimate the size of the animal's skull (and by extension, its body) when using the narrow-snouted extant Cuban crocodile as a template. They estimated the skull to be 2 m in length. Better specimens of another skull were recovered in 2002 which showed that Deinosuchus had a smaller skull than initially believed, at 1.31 m in length, and that the snout was broad and short, rather than large and narrow. This also affects the estimates of the body length. Instead of the original 15 m estimation, Deinosuchus is now believed to have been 12 m. Still large, but not monstrously so.--  Obsidi ♠ n   Soul   15:48, 26 November 2011 (UTC)

OK, than. Thank you for your answer! Regards. — Preceding unsigned comment added by 89.123.184.181 (talk) 09:07, 28 November 2011 (UTC)

inventor of potentiometer(voltage divider)
Is Pogenndorff the inventor of potentiometer(voltage divider) or potentiometer(measuring device)? — Preceding unsigned comment added by 117.206.43.166 (talk) 17:01, 26 November 2011 (UTC)
 * Johann Christian Poggendorff indicates he was involved in developing the Potentiometer (measuring instrument). However Potentiometer indicates that he invented the voltage divider instead.  However, that appears to be quite incorrect; the text at Johann Christian Poggendorff makes it clear that the device he invented was a measuring instrument, and NOT a voltage divider.  The text from Potentiometer should probably be removed.  -- Jayron  32  18:36, 26 November 2011 (UTC)
 * Note: I've gone ahead and fixed the mistaken statement, so all articles should be correct now. -- Jayron  32  05:12, 27 November 2011 (UTC)
 * The "voltage divider" would have been built and used very early on after scientists started experimenting with batteries, conductors and resistors, but did not apparently gain that name until the 1890's per a Google Book search. "Potential divider" was a term used from about 1866 on, attributed to Sir William Thompson.  The potentiometer, if one means a bridge circuit for measuring voltage without drawing current from the circuit, and using a standard cell, is a much more complex and intricate device which came along in 1841 Similar devices were still in common use in the late 20th century for calibrating other volt meters. Edison (talk) 01:25, 28 November 2011 (UTC)

"Tangential" Relativity
In pretty much all of the articles I've read on Special Relativity, it deals with the special case of standard configuration where the origins of the reference frames coincide at some point. This means of course that the relative velocity vectors have to be directed radially outward with respect to the observers. If we had a situation where the velocity was not radial but instead perpendicular to the line of observation (i.e. - tangent to an imaginary circle or orbit around the observer), would the effects of length contraction/time dilatation/etc be dependent in the same way on the magnitude of velocity as it is for standard configuration? — Trevor K. — 18:01, 26 November 2011 (UTC)  — Preceding unsigned comment added by Yakeyglee (talk • contribs)


 * The mathematics of the Lorentz transform are usually notated with vectors; our article explains several special cases, including the y and z directions. The general formula can be written as a rank-four matrix, so that you can treat all dimensions simultaneously.
 * If, as you described, the velocity of the reference-frame was "tangential" to a circle (as observed in the laboratory coordinate system), that would require a constantly-changing velocity vector as time progresses: in other words, an accelerating or rotating frame of reference. Non-inertial reference-frames require general relativity.  We have an article, non-inertial reference frame, and discussion of the implications for relativistic speed.  Nimur (talk) 18:40, 26 November 2011 (UTC)


 * Non-inertial reference frames don't require general relativity, just as non-Cartesian coordinate systems (e.g. polar coordinates) don't require non-Euclidean geometry. -- BenRG (talk) 22:32, 27 November 2011 (UTC)
 * My error. The math used to describe such a situation would be a "more generalized formulation" of the Lorentz transform (that is - more general than a one-dimensional treatment); but BenRG is correct; this scenario would not require the complete generalization of gravitation and the geometry of space and time that we usually call "general relativity."  Nimur (talk) 01:30, 29 November 2011 (UTC)
 * I can't figure out exactly what situation you're asking about, but as I said in the thread immediately below this one, all of these "phenomena" exist also in Euclidean geometry. Your question boils down to "if I have a block in such-and-such a place and rotate it in such-and-such a way, what will happen to its horizontal dimensions?" If you can manage to solve the Euclidean version with geometric intuition (or experimentation) then the answer in special relativity will be the same, except probably with the sign reversed (contraction instead of expansion). -- BenRG (talk) 22:32, 27 November 2011 (UTC)

Special Relativity: how does it work?
Presentations of special relativity state that two observers in relative motion to each other will find different values x and x' when measuring their distance to a remote object. How can this difference be due to their relative motion?

Both observers might obtain different values from similar measurement processes if exposed to different experimental conditions. Should this difference consist in their relative motion, then the velocity of each observer is objectively different, it becomes an absolute quantity, in full contradiction with the principle of relativity.

Are both observers exposed to relativistic effects? On which ground can one predict whether x will be larger or smaller than x'? Which physical parameter is the trigger? Sugdub (talk) 19:45, 26 November 2011 (UTC)
 * Two observers whose motion relative to a distant point are different velocities will measure the distance TO that point as different because of length contraction. In simpler terms, lets say that I am standing still relative to some star, and I measure the distance to the star as 100 light years.  If you are in a fast-moving space ship, and pass me, and measure the distance to that same star at the exact instant you pass me, and thus are (in my reference frame) 100 light years from the star, you will measure the distance to that star as less than 100 light years in your reference frame.  The formula at the top of the Length contraction article is how you calculate the difference between our measurements.  -- Jayron  32  19:52, 26 November 2011 (UTC)
 * And note that the differing velocities do not cause an error in measurement. The faster ship is measuring a shorter distance because the distance in its reference frame actually is shorter than the same distance (ships to star) in the slower ship's reference frame, and each reference frame is equally valid.  There is no preferred frame of reference. -- 203.82.66.204 (talk) 01:45, 27 November 2011 (UTC)


 * These answers make some good points, but I'm not sure that's exactly what the OP wants to know (perhaps Sugdub can clarify further if we're all off course). As I understand it, the question is really about the violation of symmetry that appears to take place in relative motion. My attempted explanation is this. Let's say you have quadruplets A (on earth) B (who takes off in a rocket ship from earth), C (who remains at a distant star, not moving with respect to earth) and D (in a rocket ship travelling in the same direction and with the same speed as B, some distance apart). Suppose B is en route for C at 80% of the speed of light (so I believe time goes at 0.6 the rate of earth time). The length to C is contracted in B's reference frame, so B thinks the distance is much shorter. In A's reference frame, because of B's motion, the distance between B and D is contracted. This is where the symmetry is preserved: if you are measuring the distance between two things that are co-moving (moving at the same speed in the same direction) their relative distance is what contracts relative to you. Einstein used the example of a train - read anything by Brian Greene for a very clear explanation of this thought experiment. The two ends of a train carriage are co-moving, so the distance must be contracted, so the train carriage must be shortened. In my example, the ends of the train are analogous to the people B and D. Hope I've helped instead of confusing the issue. IBE (talk) 13:02, 27 November 2011 (UTC)

All three responses tell me that the "length contraction" property/theorem is the solution... but my question deals with the feasibility to derive the "length contraction" property itself from hypotheses which only state that both observers are in relative motion to each other. It does not work... An hypothesis is missing in order to constrain the outcome of the reasoning so that x' is smaller than x better than the opposite. In order to derive the Lorentz transformation and the "length contraction" property, physicists must use an additional hypothesis which is not spelled out. As anticipated by IBE, I must say that all three responses are actually off course: none of the answers indicates which physical parameter triggers x' being smaller than x.

The "length contraction" property cannot be due to the relative motion between both observers: since it means that one observer finds a lower value than the other (thereby creating an order between them), one must rely on a non-symmetrical criterion to decide which observer will find this lower value, whereas the "relative motion" criterion coupled with the relativity principle means reciprocity and therefore does not specify any kind of objective ordering between both observers.

Reading again the input from Jayron, it appears that such a non-symmetrical hypothesis is actually involved in your reasoning, although you do not seem to be conscious about it: "… lets say that I am standing still relative to some star, and I measure the distance to the star... ". You assume being at rest in respect to the star and you imagine another observer in motion in respect to you, and this necessarily means that the second observer is not at rest in respect to the star you are both looking at. So a non-symmetrical criterion is involved in your reasoning: the relative speed between each observer and the target star. You assume that both observers have a different relative speed in respect to the object they both look at, that one moves faster than the other from/toward the star...

The input from IBE follows somehow the same path: the difference in experimental conditions is specified in terms of the relative speed (magnitude and direction) to the star. The third input insists on the equivalence of all reference frames, but does not mention the very specific role played by the reference frame of the target object, in which the first observer is assumed to be at rest …

Incidentally, due to their different relative speed to the star and assuming they operate at the same time (which is not mandatory since one of the observers is at rest in respect to the star), both observers will appear being in relative motion to each other, but this is only an epiphenomenon, a non-significant consequence of the objective difference in their respective experimental conditions. Once the relative motion of each observer in respect to the star has been properly specified, the relative motion between observers is fully determined. But this does not work in the opposite way.

Indeed this criterion, should it be clearly spelled out in terms of its magnitude and direction, has the potential to trigger which observer will find the lowest value for his/her distance to the star.

Eventually, the property of "length contraction" cannot be logically derived from the relative motion of observers. It must be traced to an objective difference in their respective experimental conditions: their different relative speed in respect to the target object. At least this is what I can conclude from your reasoning, although physicists dealing with special relativity do not seem having a clear and logical line on this issue.Sugdub (talk) 14:54, 28 November 2011 (UTC)
 * Length contraction follows immediately from the Lorentz transformation. You can derive the Lorentz transformation from two postulates: 1) The laws of physics are the same in all inertial frames of reference.  2) The speed of light is the same constant value in all inertial frames of reference.  (You actually also need to postulate the homogeneity and isotropy of space, but those are generally taken to be obvious).  For a derivation of the Lorentz transformation from those postulates, see Lorentz transformation.  Red Act (talk) 16:29, 28 November 2011 (UTC)


 * Suppose you have two rectangles in the Euclidean plane, A and B, that are the same shape but at different angles. "Relative to A", B is wider than A. But "relative to B", A is wider than B. "Relative to X" means that you cover the plane with a Cartesian coordinate grid that's aligned with the sides of rectangle X and make measurements using that. "Wider" refers to the fact that, with respect to a horizontal/vertical Cartesian grid aligned with A, the horizontal distance between the sides of B is larger, as shown in this terrible ASCII art:

____         /\      |    | A      /   \  B      |....|       /...../ |   |      /     /      |____|       \   /                     \/
 * If m is the slope of the side of B, then the width of B, relative to A, is larger by a factor of $$\sqrt{1 + m^2}$$.


 * Length contraction in special relativity is exactly the same thing, except that, because the geometry is slightly different, the width of the inclined object is smaller instead of larger, by a factor of $$\sqrt{1 - v^2}$$. There's no contradiction in saying that A is contracted "relative to B" and B is contracted "relative to A", because of what "relative to X" means here. A harder question is why "relative to X" should mean drawing a Cartesian coordinate system aligned with X in the first place. There is no good reason. It's just a convention that you will have to learn in order to pass an undergraduate special relativity course.


 * This convention is not Einstein's fault, incidentally. In his original paper, he constructed coordinate systems explicitly (from clocks and metersticks) and didn't call them "observers". He did use the word "observer" to refer to just what you'd expect: a scientist making observations. Later popularizers apparently tried to copy Einstein's language without really understanding it, and came up with the current mess.


 * Similar questions have come up before on the RD. Here are some previous threads where I ranted on this subject, perhaps better than I did this time: Time_quickens?, Speed of light and time, Speed though time. -- BenRG (talk) 04:16, 28 November 2011 (UTC)

See also here. Count Iblis (talk) 16:51, 28 November 2011 (UTC)