July 26 Information

Okay, lots of random things happened in a gas and dust cloud a over five billion years ago and as a result of these random interactions Sol system formed, along with the Earth. Or did it happen that way?

Take a random photon released from that cloud say six billion years ago. A lot of these escaped into extra-galactic space. So one fine day thirty billion years from now, some alien astronomer will capture one of these photons and measure its polarization with respect to some arbitrary angle.

What does this have to do with us? Well this photon will of course be in a state of Quantum entanglement with one atom from that primordial gas and dust cloud. So depending on which direction she decides to measure that polarization, the Earth will either have existed or not. Right? Hcobb ( talk) 01:09, 26 July 2013 (UTC) reply

How does entanglement factor into the observer's conclusion of the past existence of Earth? Plasmic Physics ( talk) 01:28, 26 July 2013 (UTC) reply

Depending on the state of that atom it would have absorbed or ignored another passing photon which would have interacted with another atom and so on. So for a lack of a nail the kingdom could be lost or in this case the Earth either formed or not. And none of this will be decided until that one measurement billions of years from now in a galaxy far away. Hcobb ( talk) 02:39, 26 July 2013 (UTC) reply

Only in the Copenhagen interpretation. — Quondum 02:48, 26 July 2013 (UTC) reply

That actually sounds like a modification of the many-worlds interpretation recently suggested, whereby the past is also indefinite as a consequence. Now this is just me, but won't there be more than one photon involved in this? You can't really infer a whole lot about a planet's (non)existence from a single photon. Plasmic Physics ( talk) 03:14, 26 July 2013 (UTC) reply

In order to make sense of this question, you have to explain why an alien astronomer looking at a photon provides a fundamentally different interaction than all the other interactions that have happened to the Earth. If we assume there is such a process as wavefunction collapse with the property that observation destroys entanglement (which isn't entirely clear), then interacting with either side of the entangled system can cause that collapse. In particular, a nearly infinite number of interactions involving the Earth could have collapsed the Earth-photon system long, long ago, and as a result the photon would already be in a determined state long before the alien astronomer ever looked at it. Hence it would appear that his interaction loses the ability to actually change anything. Dragons flight ( talk) 03:25, 26 July 2013 (UTC) reply

• If I understand you correctly Hcobb, you are stipulating that at some point before the earth existed the quantum state of various particles would, by the butterfly effect, have influenced the existence or non-existence of the earth, and that at some point in the future, one of those entangled states will be collapsed, leading to the existence or non-existence of the earth now. But the earth exists now, and the quantum state cannot have been collapsed yet, so the Copenhagen interpretation is false. You're not claiming to know which ancient photon or which future physicist. You don't believe you have have to if I understand correctly. To deny it would be claim there's something special about the earth; that no future alien could observe a "pre-earth" entangled photon quantum collapse go in the wrong direction, meaning indeterminacy is false. μηδείς ( talk) 03:57, 26 July 2013 (UTC) reply

It just doesn't work that way. It's hard to go into more detail without going into the actual mathematics, but although there is a measurement effect, it doesn't cause planets to cease to exist. That would be silly. Don't confuse the articles about "quantum weirdness" in magazines like New Scientist with reality. -- BenRG ( talk) 07:20, 26 July 2013 (UTC) reply

The Earth does not exist in some sectors of the multiverse, we happen to be in a place where it does exist. So, what happens is that this future observer can find himself in our sector or he may find himself in some other sector. In fact, both outcomes will be realized. Count Iblis ( talk) 12:35, 26 July 2013 (UTC) reply

When you observe the system, you'll either see an Earth or not, just like Schroedinger's cat. Now that darned cat, if alive, had the chance to observe that a Geiger counter failed to tick one particular second and some evil-looking phials automatically disarmed themselves; but to those outside the box, neither the cat nor what it has observed are yet sure to exist. Likewise to us the Earth is observed, but perhaps to someone else it is still in a superposition of states and might not exist at all. Now I will leave it to you to consider whether Schroedinger's cat "really" saw that Geiger counter measurement happen, before you as the conscious observer ever opened the box and determined which way it went. Wnt ( talk) 21:32, 26 July 2013 (UTC) reply

• The question is struggling to find the Anthropic principle. I suggest a reading of that. -- Jayron 32 13:29, 27 July 2013 (UTC) reply

How tall can a vertical lego structure be before it can no longer support its own weight? I'm not talking about a pyramid or something else with a fancy way to distribute the load. Really just, how tall can a stack of legos get before the bottom ones get crushed by the weight of those on top? Thanks. Someguy1221 ( talk) 01:28, 26 July 2013 (UTC) reply

Assuming that you are talking about size 1 x 2 x 4 'Lego' branded bricks, following the current manufacture specifics, it would depend on the vertical anisotropic compressibility of an individual brick, the local gravity gradient with respect to height, ambient temperature at ground level. Plasmic Physics ( talk) 01:35, 26 July 2013 (UTC) reply

Search through BBC News, I believe they had a detailed consideration of this (supported by the people from LEGO). The answer is quite a lot. -- Demiurge1000 ( talk) 01:37, 26 July 2013 (UTC) reply

According to [1] -- a third of a million bricks tall, about 2.2 miles high. SteveBaker ( talk) 01:55, 26 July 2013 (UTC) reply

Wouldn't such a stack topple over due to its high CG? 24.23.196.85 ( talk) 05:37, 26 July 2013 (UTC) reply

It is almost certain to topple over or buckle but that was not in the question. Richard Avery ( talk) 07:58, 26 July 2013 (UTC) reply

Here is a link to the excellent Dec 2012 article on BBC News Magazine. The compressive strength of a 2x2 lego brick was tested by the Open University's engineering department for the Radio 4 programme More or Less. Gandalf61 ( talk) 09:01, 26 July 2013 (UTC) reply

And, although you have the answer to your question, I have the answer for a pyramid. After those numbers were published I wrote a simple program to see how tall a tower can get if you do build out the base. Building from the top down, and only growing the bottom layer once it is too weak to support the mass above it, you can reach over 88km high, with an 11km wide base. The big assumptions in my program are that the load is distributed evenly among a layer with no uneven forces and that the surface it is built on is flat. A one-brick wide tower (like discussed above) is enough to reach geosynchronous orbit on Phobos or Deimos. 209.131.76.183 ( talk) 11:43, 26 July 2013 (UTC) reply

OMG! A working Lego space-elevator! That would be quite a neat project! :-) SteveBaker ( talk) 19:43, 26 July 2013 (UTC) reply

There are no geostationary Earth orbits possible on Phobos or Deimos. DreadRed ( talk) 20:51, 26 July 2013 (UTC) reply

My program only knows the mass, radius and rotation period of a body - it doesn't know Mars is be there to interfere with things or that the moons aren't very regularly shaped, or anything else that happens to prevent that orbit. :-) It would happily tell you how to build a tower on a spherical cow if you asked it to. And Steve, that's exactly why I wrote the program - after I linked an article on how high a 1-brick-wide tower could get to a friend, he told me that the real question is if you could build a space elevator with them. The structure also had to be entirely compressive, since glue is not allowed! 209.131.76.183 ( talk) 11:51, 29 July 2013 (UTC) reply

Since events like weddings are often planned months in advance, and there is an expectation for the nuptials to be consummated on the night of the ceremony, how have women made sure that they won't be on their period on that night? I know it's possible to gain or lose a few days either side with contraceptive pills, but how was this accomplished in times or in places where it's not available? 202.155.85.18 ( talk) 04:11, 26 July 2013 (UTC) reply

Comments retracted. Vespine ( talk) 04:37, 26 July 2013 (UTC) reply

The question makes outdated sexist assumptions about a wife's duty to perform sex. DreadRed ( talk) 21:04, 26 July 2013 (UTC) reply

Nonsense. The question makes no such assumption, and many women prefer not to have sex while menstruating, but look forward to it on their wedding night. It's almost like someone is assuming sex is bad for women. μηδείς ( talk) 21:11, 26 July 2013 (UTC) reply

The substitution "there is an expectation" for "there used to be an expectation" is provocative. While the latter is consistent with 19th century mores the former is not a dictat to which people today are obliged to submit. The claim concerns the presentability of the bride for penetrative coitus i.e. sexual intercourse, at the predefined time and place. The euphemism "nuptuals[sic] to be consummated" casts an outdated prurience on the subject; to my knowledge "nuptual" is an adjective, not a noun. The OP neither assumes sex is good nor bad per se for women but does assume that unsuitable preparation of the bride's genitalia would breach society's expectations of her. What is bad for women, and need no longer be their lot, is their diagnosis as hysterical by ignorant physicians est femineo generi pars una uterus omnium morborum "the womb is a part of every illness of the female sex" and the cognitive dissonance in male dominated religions that regard menstruation as divinely ordained yet so unclean as to be unmentionable. Medieval Catholicism stipulated that menstruating women were not to join ordinary sinners in church, so the OP's question is applicable at least in that outdated context. The famed wedding night accomplishment may not be as important as the OP's question and romantic fiction suggest because if the groom is ignorant of a woman's menstrual cycle, he will get that news in the weeks after if not actually on the wedding night. The shock should be tolerable. But anyone interested in joining speculations why the expected was not accomplished on a Victorian wedding night 10 April 1848 can read about Effie Gray and John Ruskin. DreadRed ( talk) 10:31, 29 July 2013 (UTC) reply

Wow. Thanks for that wall of nonsense. Reading a bunch of perceived slights into my question might entertain you, but it sure isn't consistent with assuming good faith. 202.155.85.18 ( talk) 08:48, 30 July 2013 (UTC) reply

I'm not sure, but I presume that the bride can chart her cycle and use that to plan her wedding day? (Of course, I'm not an expert -- I'm a man, and single!) 24.23.196.85 ( talk) 05:35, 26 July 2013 (UTC) reply

She could estimate it pretty closely, assuming her monthlies are consistent; and if she's on the pill she can figure it out to the day. ← Baseball Bugs ^{What's up, Doc?} carrots→ 11:50, 26 July 2013 (UTC) reply

Not a direct answer, but in Victorian times in England at least, many weddings took place over Christmas or Easter holidays so people didn't lose time off work. If a woman's "friend was visiting" at those times, well either they got on with it or waited. -- TammyMoet ( talk) 13:13, 26 July 2013 (UTC) reply

I've heard of doctors prescribing birth control pills (or possibly "day after" pills) to induce an early menstrual cycle a couple of weeks before some important event - but that's getting into the realms of medical advice (which we're not allowed to give) - and you should definitely consult a doctor before considering something like that. SteveBaker ( talk) 19:41, 26 July 2013 (UTC) reply

Can the Skip Doctor be safely used not to repair a scratched CD, but to clean a CD that's merely dirty rather than scratched? 24.23.196.85 ( talk) 05:32, 26 July 2013 (UTC) reply

I wouldn't. The way data is stored on a CD to avoid errors is to store error correction bits scattered around one circular track of the disk. This is done so that if a scratch goes at some random angle across the disk, it's unlikely to take out both the data itself and the error correction information that is there to help to recover from scratches, etc. If you happen to get a bit of grit or something when wiping or rubbing in a circle - it causes a mark that hits all of the error correction bits of one track...bad news! However, if you wipe or rub in a radial fashion (from the middle to the outside or outside to the middle) and something goes wrong, you'll only take out a small section of each track - which the error correction software will happily deal with - so you won't hear a glitch. As I recall, the "Skip Doctor" inherently does circular wipes - which is a bad idea. Use it if you must to fix bad scratches when all else fails - but not for routine cleaning.

I suggest following this "instructible" which seems to have the right idea.

(You might like to know that I was one of the team at Philips Research Labs that made the very first CD-ROM - so I have some idea of what I'm talking about!)

SteveBaker ( talk) 19:37, 26 July 2013 (UTC) reply

A video disc worked on by Compaan and Pete Kramer at Philips was released in 1978 but failed to become a standard. The use in CD-ROMs of 16-bit audio and Reed-Solomon forward error correction were both proposed by Sony to Philips in 1979 [2]. The SkipDr(c) device comprises a buffing wheel whose axis is perpendicular to the CD axis. It can be seen here to stroke the disc radially, not circularly, which is consistent with the advice on routine cleaning. DreadRed ( talk) 20:32, 26 July 2013 (UTC) reply

I was working at Philips straight out of college in 1977. The first commercial CD audio (not CD-ROM) drives were being prepared for sale in around 1980 to 1981 based around the SONY proposal from 1979 - that specification didn't include computer data, it was just an audio spec. We took a pre-production audio drive and modified it to work with our prototype "CHRIS" home computer - we made an interactive dictionary application, but because we didn't have time to type in all of the data, paint the pictures and record the audio - we only did the letter "O". My part was to write a "paint" program for the CHRIS computer to allow an artist to paint pictures for our prototype application. We had 50 disks pressed at the Philips CD factory - which was ramping up to make the first audio CD's. The biggest problem was that affordable hard drives topped out at 20Mbytes at the time - so authoring 720Mbytes of data to fill a CD-ROM was a serious problem! We demoed the system to hundreds of people in the industry - including computer manufacturers - but sadly, we didn't consider the disks to be very important and nobody knows what eventually became of them - not one of them is known to exist. Philips sold a batch of their first generation CD audio players in the staff shop several months before they appeared on the open market in order to see how they fared with "real" users. So I owned a CD player before they were ever released to the public. Sadly, the staff shop only had 3 CD's on sale...some stuff by Bach, a digital re-mastering of various Glen Miller tracks that was taken from the original wire recordings and Dire Straits "Brothers in Arms". All three CD's still play well.

Anyway, I guess I was thinking of some other device than the Skip Doctor. If it strokes radially then it should be OK - I thought it worked in a circular manner, which would be bad.

SteveBaker ( talk) 01:51, 27 July 2013 (UTC) reply

Why do malignant growths tend to have a more deformed appearance compared to benign growths? I'm not sure if "deformed" is the right word but benign growths tend to look less "messed up" to use a less scientific word. Clover345 ( talk) 11:46, 26 July 2013 (UTC) reply

What makes them malignant is that they do tend to invade other tissues. So it's (mostly) a tautology. -- Stephan Schulz ( talk) 14:25, 26 July 2013 (UTC) reply

Since we're having great fun with quantum mechanical thought experiments, let me post a problem that you can actually do some computations about. A person enters a perfectly isolated box and is trapped in there for a thousand years. After a thousand year the box is opened. What is the probability that the person has not aged at all? Count Iblis ( talk) 12:46, 26 July 2013 (UTC) reply

Zero. Next question? AndyTheGrump ( talk) 13:12, 26 July 2013 (UTC) reply

Count Iblis is probably alluding to the very same "piano in a box" paper he's posted a few times before, which is a very interesting read. There's a finite but non-zero chance the box's contents will be exactly as they were before the box was closed, after random-walking its way through the space of all possible states. But for most purposes, we can consider that probability to be so small as to be indistinguishable from zero by any physical measurement; and that's the end-all for observational science. Nimur ( talk) 15:03, 26 July 2013 (UTC) reply

You can't have a perfectly isolated box that you can still open in a thousand years. Once it's perfectly isolated (e.g. you drop it into a black hole) then it is causally separated from your part of the universe and you cannot open it. At best you could put a timer on the box that automatically opens it after a set amount of time has elapsed (in the box's proper time) but then you can't observe the box when it opens anyway. So this question goes in the same file as "what happens when you apply an irresistable force to an immovable object ?". Gandalf61 ( talk) 15:32, 26 July 2013 (UTC) reply

Just take some arbitrary box, the answer will then depend on the assumptions made about this (consider e.g. the well known extreme case of continuous measurement, of course this isn't realistic either). Count Iblis ( talk) 19:40, 26 July 2013 (UTC) reply

The molecules in the box might somehow (through random brownian motion or whatever) wind up in the exact same state as they were the instant the box was closed...so I guess there is a vanishingly small - but non-zero - chance that the person would emerge imagining that no time had elapsed. He has a much much better chance of emerging as a 150lb unicorn who speaks only fluent esperanto, with a bunch of tulips for a tail, hooves made of chocolate and ears that sing La Traviata in perfect harmony...but yep, it's possible. You don't even have to wait a thousand years - a couple of minutes is plenty if your tolerance for extreme improbability is good enough. SteveBaker ( talk) 00:56, 27 July 2013 (UTC) reply

Nimur actually answered this better above: the difference between calculably infinitesimal and measurably impossible converge to zero for problems like this. That is, this kind of situations demonstrates the problems with paper physics: you can create all sorts of situations on paper which you cannot make the math prove are actually impossible to an infinite number of decimal places; but we forget that the math is merely a model of reality, and in the limits we're talking about, we're well below the tolerances of the physical world, tolerances that don't exist on paper because we can arbitrarily be as infinitely accurate as we want in the math. Nimur's statement "But for most purposes, we can consider that probability to be so small as to be indistinguishable from zero by any physical measurement; and that's the end-all for observational science" is sufficient to end the silliness of questions like this. -- Jayron 32 14:38, 27 July 2013 (UTC) reply

Nimur is wrong here. When the underlying science is well established you can look into these sorts of questions in a meaningful way. Observational science has already fixed so much of the foundations that you can't invoke the fact that you can't actually measure this particular effect as an argument why this isn't meaningful science.

A good example of how this sort of criticism goes wrong is the way the Maxwell's demon though experiment was treated. It leads to a new insight about entropy, basically that the information theoretical definition of entropy is fundamental to the thermodynamic definition, you can't escape the information theoretical definition even if you are doing down to earth thermodynamics. However it took almost century for this insight to arise, the standard treatment of Maxwell's demon was to show how some particualar implementation of the thought experiment would not work. But that was besides the point, the point is that in theory you can come up with an implementation that will work, albeit ever so slightly which then forces one toward the information theoretical definition of entropy. Count Iblis ( talk) 12:25, 28 July 2013 (UTC) reply

So, this problem points to time evolution being ambiguous precisely due to the time evolution of the wavefunction. But if you consider time evolution more deeply it gets even more ambiguous, as pointed out in this paper. Count Iblis ( talk) 14:55, 30 July 2013 (UTC) reply

I'm not sure if this has been asked before, but here it goes. Take this scenario: you are inside a spaceship which is travelling faster than the speed of light. You look through a window. Theoretically, based on the current understandings in physics, what would the outside look to your eyes: that is, what will you see? Yes I know there's a theory that nothing can go faster than the speed of light. The scenario is only theoretical. Narutolovehinata5 ^{t c csd new} 13:57, 26 July 2013 (UTC) reply

White I guess? ☯ Bonkers The Clown \(^_^)/ Nonsensical Babble ☯ 14:02, 26 July 2013 (UTC) reply

It a theory says you can't do something, it can't tell you what would happen if you did. AndyTheGrump ( talk) 14:04, 26 July 2013 (UTC) reply

Actually, the Theory of Relativity, as far a I understand it, does not say that you cannot travel faster than the speed of light. It only says that you cannot accelerate anything with a non-zero rest mass to the speed of light, and that any kind of FTL signal propagation throws causality out of the window. Only if you assume that going to FTL implies going at the speed of light in between, or if you assume causality as a given (which many people do implicitly, and some explicitly) does FTL travel become strictly impossible with the ToR. -- Stephan Schulz ( talk) 14:15, 26 July 2013 (UTC) reply

If you are traveling faster than the speed of light away from the objects you are looking at then the light emitted by or reflected by those objects would not reach your eyes, so you would not see them, it would be black. 50.101.146.152 ( talk) 14:09, 26 July 2013 (UTC) reply

It would be like that bit in 2001 Space Odyssey just before he ends up in the white room. Zzubnik ( talk) 15:02, 26 July 2013 (UTC) reply

No, that effect was produced by lighting and camera tricks, including slit-scan photography. That sort of neat-looking geometric distortion doesn't require faster-than-light speeds; it's actually a very simple mechanical camera trick that has been used by so many filmmakers that it's become part of the collective cultural interpretation of "hyper speed." Nimur ( talk) 15:16, 26 July 2013 (UTC) reply

I don't know where this business of "Physics doesn't say that you can't travel faster than light" comes from. Physics most definitely does say exactly that. Plug a velocity greater than 'c' into any of the relativistic equations and you'll soon find that you'll need to calculate something like this (this happens to be the equation for time dilation - but it's the same thing for mass, length, etc):

${\displaystyle \Delta t'={\frac {\Delta t}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$

If v (your velocity) is bigger than c (the speed of light) then v squared over c squared is bigger than 1. One minus that is therefore a negative number - and you need to take the square root of that. Taking the square root of a negative number is decidedly problematic. The answer is either "You Can't Do That" (meaning that the equations predict that you cannot move faster than the speed of light) or that the answer is a complex number. Sadly, there has never been a real-world situation where a simple quantity like length, time or mass came out to be a complex number. Complex numbers simply don't appear in the answers of any physical systems unless you plugged a complex number into one of the parameters. Having your mass be a complex number simply doesn't have meaning...ditto time, mass, etc. So it's really very clear that the math is telling us that nothing goes faster than light.

Any speculation on what would happen if you could is nothing but random speculation - it's utterly meaningless and without basis in physics. Before you can do that, tell me what a complex-number length means.

SteveBaker ( talk) 18:46, 26 July 2013 (UTC) reply

I don't agree at all. There are solutions of the relativistic equations that allow for FTL, see e.g. tachyon (which is not just Star Trek techobabble). Imaginary numbers occur quite often in physics - ${\displaystyle f(x)=ae^{it}+b}$ is a standard ansatz for many differential equations. -- Stephan Schulz ( talk) 19:27, 26 July 2013 (UTC) reply

Well, to play Steve's advocate here, he didn't say the relativistic equations don't allow FTL; he said physics doesn't. That does seem to be true, provisionally, given the current state of knowledge, until someone falsifies it. -- Trovatore ( talk) 19:30, 26 July 2013 (UTC) reply

Oh, wait a minute — maybe he did say that. I think I was thinking of the section below. -- Trovatore ( talk) 19:34, 26 July 2013 (UTC) reply

Steve's mathematical expression is an algebraic restatement of the widely-known consequence: for any object traveling faster than light, then there are certain reference-frames where its behavior is anticausal. This is also called a paradox of the " tachyonic antitelephone." Strictly, we can't say this physical phenomenon is impossible through inspection of the equation. We can use the equation, though: we use its mathematical formalism to show that a physical behavior (faster than light travel) is strictly equivalent to another physical behavior that is never observed in the real world (anti-causal behavior). We do not ever see an event A causing another event B to take place prior to event A. Therefore, we conclude from observation that there are no scenarios where anti-causal behavior exists; or, because it is a mathematically equivalent statement, we conclude that no physical object travels faster than light. For the ultra-pedantic: this strict equivalence between causality and time-reversal only applies to events that interact (cause each other) through electromagnetic means. In principle, we would need an even stronger formalism to prove that there exists no fundamental interaction mechanisms that occurs faster than the electroweak interaction. We have directly observed four fundamental interactions; by definition, these account for all known interactions; and we know of none whose time-constant is faster than electromagnetic interaction. Nimur ( talk) 21:15, 26 July 2013 (UTC) reply

I don't know what you are talking about! I'll go to sleep in an hour or so, which even now makes me tired! I'm drinking some tea now, which made me thirsty! And I mentioned the causality assumption above because you did now! ;-) -- Stephan Schulz ( talk) 21:25, 26 July 2013 (UTC) reply

It seems that my most recent post caused quite a bit of confusion. I apologize. Nimur ( talk) 21:32, 26 July 2013 (UTC) reply

We have ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2, so an imaginary time interval will appear to be a spatial distance and imaginary distance will appear to be a time interval. Count Iblis ( talk) 19:34, 26 July 2013 (UTC) reply

And how do you plan to wiggle out of how mass is altered by relativity? Complex masses? Mass becomes spatial distance and vice versa? Ow! SteveBaker ( talk) 00:48, 27 July 2013 (UTC) reply

I don't see that there's anything you especially have to wriggle out of. The Lorentz transformation equations tell you how to translate between different inertial frames, and yes, if you naively plug in v greater than c, you get imaginary answers. But who says a tachyon has to have a frame of reference? Maybe the answer to the question "what would I see if I were riding on a tachyon?" is just "you can't ride on a tachyon, but that doesn't mean they don't exist". It could be that tachyons are just a different sort of beast, and there is no frame of reference associated to them. Likewise, maybe they don't have an imaginary rest mass so much as the concept of rest mass just doesn't apply to them.

Now, as far as I know, there's no evidence for the existence of tachyons, and maybe it's arguable that they're inconsistent with particle physics as currently understood. But even if so, that's particle physics, not relativity. Relativity by itself does not exclude them. -- Trovatore ( talk) 00:58, 27 July 2013 (UTC) reply

a tachyons three dimensions could be past present and future. Without moving at all in our 3d space. 68.36.148.100 ( talk) 02:24, 27 July 2013 (UTC) reply

As far as I can tell, that makes no sense at all. -- Trovatore ( talk) 03:12, 27 July 2013 (UTC) reply

Seconded! -- Stephan Schulz ( talk) 07:10, 27 July 2013 (UTC) reply

Steve, here is a 1967 paper that does in fact assume that tachyons have an imaginary rest mass. As it says, such an assumption is only a physical problem if the rest mass is observable, which need not be the case if tachyons can not ever be slowed to move at less than the speed of light. Dragons flight ( talk) 17:09, 27 July 2013 (UTC) reply

you dont need spaceship , you can see da ja vu at your home , thanks water nosfim — Preceding unsigned comment added by 81.218.91.170 ( talk) 14:26, 27 July 2013 (UTC) reply

When one reaches the speed of light, the distance between where one left and where one is going is reduced to zero. In other words, one reaches one's destination instantaneously. Travelling faster is inconceivable. TFD ( talk) 00:15, 29 July 2013 (UTC) reply

Not exactly - travelling at the theoretical speed of light is a "singularity" and mathematical functions which reach a singularity can, indeed, exist on the other side of it. More interesting is the theory that the speed of light has not been a universal constant at all -- that it specifically varied at the time of the Big Bang, with a number of distinct theories using it to explain some interesting observations. Collect ( talk) 00:28, 29 July 2013 (UTC) reply

Kinda related to a previous question where someone correctly pointed out that Special Relativity doesn't explicitly forbid faster than light travel, but states that it's impossible for an object with a rest mass above zero to accelerate to the speed of light, but something already moving FTL doesn't violate Special Relativity as long as it didn't accelerate to it. Of course, light has a rest mass of 0, so... let's say you have a positron, and an electron, you annihilate them together and you get two gamma ray photons, do those photons ever accelerate or do they just automatically move at the speed of light right away? I'm pretty sure the answer is that they don't accelerate, but I just want to be sure. ScienceApe ( talk) 15:23, 26 July 2013 (UTC) reply

When we study how light travels, we find that the best way to model it is as a pair of coupled fields - an electric field and a magnetic field. These fields interact in a way that allows a disturbance to propagate. It is a fundamental property of the universe that the rate of propagation - the speed that the disturbance moves, or the rate of change in space versus time - is always exactly constant. So, when we approximate this wave behavior by describing it as a moving particle - a photon - we say the photon has a perfectly-fixed velocity, that is always exactly c and never changes.

We can throw a wrench into the works by discussing materials with a refractive index, because analyzing the way light actually interacts with a material on a microscopic level is actually quite complicated; but for now let's stick to propagation in a vacuum. Nimur ( talk) 15:38, 26 July 2013 (UTC) reply

I've added something to the previous answer - go read that. Physics does forbid faster than light travel. Period.

As to what happens if a positron and electron collide...you don't need anything that complicated. Take a handy flashlight, press the "ON" button - note that the room lights up a handful of nanoseconds later. Of course the photons move off at the speed of light! Duh!

If you insist on a more physics'y answer - the mass of the positron and electron are non-zero. When they annihilate, that energy has to go someplace. If it's all turned into photons, then they can't be stationary because their total rest-mass would be zero and the energy would have simply vanished. The only way to get zero rest-mass to be non-zero is to move the particle at the speed of light - and that's how fast the photons have to be moving to carry away the energy from the original particles. SteveBaker ( talk) 18:53, 26 July 2013 (UTC) reply

It's not really true that physics forbids FTL travel, is it? Physics may have discovered that it's impossible, at least in the model physics uses. I can't help thinking of some scientist from Galileo's and Copernicus's time, saying "These upstarts want us to believe that the Earth and the planets revolve around the Sun, but we who know better FORBID the Sun to be there". -- Jack of Oz ^{[pleasantries]} 23:19, 26 July 2013 (UTC) reply

I'll certainly grant's that it's possible (as was the case with the Catholic Church in Galileo's day) for the current state of physics to be incorrect.

But that's a lot different from asking whether the current state of physics knowledge allows FTL travel. The current state of physics quite clearly says that FTL is impossible. We could be wrong...sure. But if you're asking "If current understanding is incorrect - is <whatever> possible" - then there is very little point in asking the question! Do unicorns exist if our knowledge of zoology is incorrect? Can you run your car on water if our knowledge of the laws of thermodynamics is wrong? Do fairies power our cellphones if electrical theory is all wrong? Who knows?! So I'm going to boldly assert that the question pertains to our current level of understanding - and that is absolutely that FTL travel results in mathematical impossibility - which means that it's impossible.

I'd go one step further. The Catholic Church believed in geocentricism for no really good reason. If you asked them why they believed that, they'd point to a vaguely worded assertion in a book of unknown age and authorship. With modern physics, we can point to a raft of evidence that shows that the mathematics of relativity fits every one of thousands of experimental observations - every single test we've ever used to test it said "Yep! Those equations fit reality."...and the very form of those mathematics is very, very clear about the impossibility of anything moving faster than light. You don't get any kind of meaningful answer when you try to plug v>c into any of the equations. You can hardly compare the state of knowledge of where the center of the earth was in Galileo's time with the piles of evidence we currently have for relativity.

SteveBaker ( talk) 00:44, 27 July 2013 (UTC) reply

Oh, but I can and I do. To be clear, I am not for a moment disputing what relativity says, nor am I taking issue with any of the experiments that have been done to test it, none of which have found it wanting. My issue is with the language you're using ("forbid"). When you go from "Let us explore and see what we find, and add our discoveries to the treasury of science" on the one hand, to "We who speak for science know the truth, and anything that diverges from it is in error" on the other hand, then you're putting yourself in exactly the same intellectual position as the RC Church of Galileo's day. Just because no educated person disputes things like relativity, gravity, yada yada, does not entitle scientists, of all people, to have closed minds, but when you forbid things to be any way but what you believe to be the case, that's what you seem to be doing. Einstein himself said "Reality is an illusion, albeit an extraordinarily persistent one" (or similar). -- Jack of Oz ^{[pleasantries]} 01:09, 27 July 2013 (UTC) reply

Steve, considering that the standard model blows up with singularities and multiverses, I wouldn't bet the farm on it. - Modocc ( talk) 06:28, 27 July 2013 (UTC) reply

Steve, I think the question is: What happens at the instant the photon is created? That is, at one point in time you have your positron and electron. In another point in time you have a photon traveling at the speed of light. If you reduce the time interval between those two points to the shortest one possible, what happens in between? (That is, what happens between the last instant you have have a separate positron and an electron, and the first instant you have a photon traveling at the speed of light? Or between the last instant you have an electron traveling in the semiconductor of an LED flashlight and the first instant you have a visible photon traveling the speed of light.) Does the photon spring into existence traveling at the speed of light, or does it somehow start "at rest" and then (very rapidly) accelerate up to the speed of light? - My understanding is that the fraction of an instant we're talking about here is so short, we're deep in the quantum realm, and talking definitively about something's existence, let alone the precise speed something is traveling, gets to be a bit problematic. If you take the sum over histories approach, as Feynman does in QED: The Strange Theory of Light and Matter (a book I highly recommend), for short periods of time and short distances the photon doesn't have a particular speed. Instead it takes all speeds, including ones which are both faster and slower than the speed of light. It's only when you go to larger distances and longer timescales (which are still microscopic to a human perspective) does the probability for a photon to be seen to be traveling at not the speed of light (either faster or slower) drop off to near zero. Additionally, for the positron+electron->photon transition, the transition doesn't happen at any one particular time or point in space, but simultaneously at a large number of times and points in space. It's also not just a simple positron+electron->photon reaction, but also happens with multiple other intermediate particles. In the path integral formulation the observed behavior is the result of all of these things happening simultaneously, with increasingly smaller contributions as things get more complicated. - At long timescales and large distances things calm down to classical behavior (photons traveling at c), but for very short timescales (e.g. the instant a photon is created) quantum mechanics behaves very strangely. -- 205.175.124.72 ( talk) 01:59, 27 July 2013 (UTC) reply

At some point someone ought to work through the Relativistic Doppler effect math - my impression is that this is not an "infinite blueshift" but, well, a negative frequency. Wnt ( talk) 01:41, 27 July 2013 (UTC) reply

The "monkey wrench" that Nimur mentions is important, for matter waves do slow down in mediums and photons are matter waves. But its more complicated than that even. For instance, its my recollection that, statistically, most x-rays propagate slower through denser materials, but a small fraction can be unaffected and travel with the vacuum speed. Its reasonable to conclude from this that this happens because some of these highly energetic photons do not interact with the material's atoms, so the intervening space between the atoms does not impede them. Essentially, condensed matter, unlike light in a vacuum, is a superimposition of various interacting matter waves which are accelerated and decelerated. - Modocc ( talk) 06:26, 27 July 2013 (UTC) reply

The invasive species Heracleum mantegazzianum is native to Caucasia and Central Asia. What are its natural enemies in those places?
— Wavelength ( talk) 15:35, 26 July 2013 (UTC) reply

People who refuse to do other people's homework for them.  ;-) 24.23.196.85 ( talk) 02:26, 27 July 2013 (UTC) reply

My question is not a homework question.

— Wavelength ( talk) 03:04, 27 July 2013 (UTC) reply

"Natural enemy" is actually a great keyword. I googled that, and the name, and found this nice paper [3], which discusses possible biological control agents, including fungal pathogens and some herbivores. SemanticMantis ( talk) 14:08, 27 July 2013 (UTC) reply

Thank you for that link. I have added it to the list of external links. (Incidentally, I noticed on the map on page 150 that one star symbol is very close to Sochi, the site of the 2014 Winter Olympics.)

— Wavelength ( talk) 16:28, 27 July 2013 (UTC) reply

Ethical issues aside, was something of value discovered? Something that we wouldn't be able to discover without breaking ethical constrains. OsmanRF34 ( talk) 17:52, 26 July 2013 (UTC) reply

It's generally agreed that the correct answer to your question is no. Not merely criminal and immoral, Nazis were also very inept scientists. They asked preposterous questions, and collected bad data. - Nunh-huh 18:29, 26 July 2013 (UTC) reply

In many cases they knew the answer that was required of them in advance and simply doctored the results to come out that way. I agree - I've never heard of anything useful that came out of that. SteveBaker ( talk) 18:33, 26 July 2013 (UTC) reply

According to the article ( Nazi human experimentation#Modern ethical issues), "Contemporary knowledge concerning the manner in which the human body reacts to freezing is based almost exclusively on these Nazi experiments." This is an issue which arises in the immediate treatment of hypothermia (for example, at sea after prolonged immersion). Tevildo ( talk) 18:44, 26 July 2013 (UTC) reply

Yes, but is that useful? Every one knows that freezing kills and that you have to rewarm victims (and maybe the Nazis discovered the exact temperature), but will this information be valuable to save a human life? OsmanRF34 ( talk) 19:02, 26 July 2013 (UTC) reply

If I recall correctly, they did discover rewarming injury, and that could be very useful. There seems to be a thread of thought, one I find utterly indefensible, that the evil undeniably employed in finding the information somehow permanently taints the information itself. That's total nonsense. -- Trovatore ( talk) 19:21, 26 July 2013 (UTC) reply

Well, there's still the second condition of my question: "we wouldn't be able to discover without breaking ethical constrains." Couldn't we have discovered that by analyzing protocols of emergency services? Couldn't we have connected the dots extrapolating from known cases? OsmanRF34 ( talk) 19:28, 26 July 2013 (UTC) reply

Oh, I wasn't responding to that. I can certainly imagine that all of this might have been found out ethically. I just have a deeply negative reaction to squeamishness about using the information, to the extent that it's useful, on the grounds that it was discovered in an evil manner ("unethically" seems far too weak a word). -- Trovatore ( talk) 19:33, 26 July 2013 (UTC) reply

The Hastings Center Report > Dec., 1984 > Should the Nazi Research be Cited? states that Nazi research data has been used in "at least 45 research articles published since World War II". Alansplodge ( talk) 19:49, 26 July 2013 (UTC) reply

It is impossible to prove that something couldn't have been discovered using ethical means. The small very sample size we have on hypothermia, outside of Nazi experiments, could be remedied by waiting several centuries for more data to accumulate. With time, somebody will be accidentally exposed to phosgene gas, and scientists will be able to deduce the gas's physiological effects. In several decades, computers may become powerful enough to simulate the entire human body, making experimentation on living things completely obsolete. The only thing we can say is that the Nazis obtained data which would be extremely difficult to obtain in other ways, because there simply hasn't been another situation where humans were frozen in a controlled manner while their physiological responses were carefully monitored. The data obtained may be reliable or unreliable, but it's undeniable that if the data is reliable, it would be the best data we have on hypothermia.

I also wonder what you consider to be "ethical constraints". The only practical alternative to experimenting on humans in experimenting on animals, and deliberately freezing animals to death isn't exactly ethical. You could make the argument that animals are subhuman, uncivilized, etc. and their suffering therefore doesn't matter, but the Nazis applied the same argument to the mentally handicapped, with the same degree of validity. -- Bowlhover ( talk) 23:24, 26 July 2013 (UTC) reply

Mmmmm, you must be one of those crazy PETA or Animal Liberation Front members. Anyway. I gave alternatives above: medical protocols, extrapolating data or simply working within a value known to be secure for humans. That means, remaining ignorant about the true safe exposure to low temperatures. For all practical purposes, you know that some value is secure, so we keep thing on the safe side. OsmanRF34 ( talk) 00:06, 27 July 2013 (UTC) reply

So? Animals are not people, so making them suffer during animal experimentation is therefore a LESSER wrong than doing this to people -- especially when the knowledge gained through animal experiments can save people's lives, in which case it's arguably a greater wrong NOT to do these experiments! 24.23.196.85 ( talk) 01:41, 27 July 2013 (UTC) reply

Yes, animals are not people, in the same way that Jews are not Aryans, or the mentally handicapped are not the mentally fit. All three statements have the same degree of moral relevance.

Anyhow, one of the reasons that the Nazi data on hypothermia is useful (if it is reliable) is that animal experiments do not, in fact, give an accurate picture of human hypothermia. This source, quoted by our article on Nazi experimentation, claims: "Animals and humans differ widely in their physiological response to cold. Accordingly, hypothermia research is uniquely dependent on human test subjects." The same article describes one of the Nazis' discoveries:

"The Nazis attempted rewarming the frozen victims. Doctor Rascher did, in fact, discover an innovative "Rapid Active Rewarming" technique in resuscitating the frozen victims. This technique completely contradicted the popularly accepted method of slow passive rewarming. Rascher found his active rewarming in hot liquids to be the most efficient means of revival."

Because hypothermia affects animals very differently, and because it kills comparatively few people (compared to cancer, for example), Institutional Animal Care and Use Committees are unlikely to approve animal experiments, especially if the animals are not tranquilized or heavily anesthetized (both of which make animal studies even less representative). Contrary to Osman's suggestion that I'm somehow an extremist, it is standard procedure for IACUCs to minimize pain and reject proposals that amount to torturing animals. -- Bowlhover ( talk) 06:42, 27 July 2013 (UTC) reply

Jews are people, retards are people, but animals are not people -- therefore there's NO moral equivalency between experimenting on animals and experimenting on the other two categories you mentioned! But you do have a point that animal experiments dealing with hypothermia are likely to give invalid data. 24.23.196.85 ( talk) 07:56, 28 July 2013 (UTC) reply

Given a sphere, i.e. the Earth, with a defined North-South axis, I would like to consider the following problem.

Consider two points on the surface of the Earth, ${\displaystyle {\vec {a}}}$ and ${\displaystyle {\vec {b}}}$, I want to define a function ${\displaystyle \theta ({\vec {a}},{\vec {b}})}$ that expresses the notion of whether ${\displaystyle {\vec {a}}}$ and ${\displaystyle {\vec {b}}}$ are aligned in a North-South orientation, an East-West orientation, or something in between. I would think the natural way to do this is to think of ${\displaystyle \theta }$ as an angle, where 0 suggests that ${\displaystyle {\vec {a}}}$ is directly North of ${\displaystyle {\vec {b}}}$, pi/2 radians (90 degrees) suggests ${\displaystyle {\vec {a}}}$ is directly East of ${\displaystyle {\vec {b}}}$, pi radians (180 degrees) suggests ${\displaystyle {\vec {a}}}$ is directly South of ${\displaystyle {\vec {b}}}$, etc.

For my application it is necessary that ${\displaystyle \theta ({\vec {a}},{\vec {b}})}$ always differ from ${\displaystyle \theta ({\vec {b}},{\vec {a}})}$ by pi radians (180 degrees) in all non-degenerate cases (i.e. ${\displaystyle {\vec {a}}\neq {\vec {b}}}$). Aside from the immediate neighborhood of the degenerate points, I also need the function to be smooth.

Maybe this seems simple, but there seem to be problems with points near the pole and opposing points near the equator (e.g. where the distance traveled over the North pole would be the same as the distance traveled around the equator).

Is there a natural way of defining such a function ${\displaystyle \theta ({\vec {a}},{\vec {b}})}$? Or is it the case that you just can't do this smoothly on a sphere? Dragons flight ( talk) 22:24, 26 July 2013 (UTC) reply

You may do better with this on the math desk. μηδείς ( talk) 22:59, 26 July 2013 (UTC) reply

It depends on how much flexibility you have in defining the function. You seem to be expecting a real-valued function that gives you an angle in radians. If you are going to constrain it this much, then the answer is clearly that you cannot do it. Even if you do not insist that it be an actual angle, but merely a real-valued function that must increase by one unit when the one point circumnavigates the other, topologically you are assured of at least two discontinuities ( Hairy ball theorem may be of interest). However, if you wish to represent a more general concept of "how must I rotate point ${\displaystyle {\vec {a}}}$ around the sphere to turn it into point ${\displaystyle {\vec {b}}}$ along a great circle", effectively simultaneously giving the axis and angle of rotation, this can be done in a globally smooth fashion, including the sticky cases of no rotation (${\displaystyle {\vec {a}}={\vec {b}}}$) and the antipodes (${\displaystyle {\vec {a}}=-{\vec {b}}}$). Your function must produce a four-dimensional vector (but with fewer degrees of freedom). If it is numerical stability you're looking for, this might not be too much of a price to pay. Related articles include Rotation (mathematics), Rotor (mathematics), Geometric algebra, Plane of rotation, Rodrigues' rotation formula. They'll be a bit much to go through in one sitting and extract exactly what you want, but if this approach makes sense to you, I can try to distil out the relevant formulae for you. — Quondum 00:20, 27 July 2013 (UTC) reply

I probably should have been more careful to note that I don't care about the coordinate cut at 0 = 2*pi. My real, i.e. underlying problem, is something like this. Given functions ${\displaystyle f_{NS}(|{\vec {a}}-{\vec {b}}|)}$ and ${\displaystyle f_{EW}(|{\vec {a}}-{\vec {b}}|)}$ meant to describe respectively the North-South and East-West components of a field that only depends on distance and orientation, is their a natural way to extend this to the whole sphere. I was imagining something like:

${\displaystyle f({\vec {a}},{\vec {b}})=f_{NS}(|{\vec {a}}-{\vec {b}}|)\times cos(\theta ({\vec {a}},{\vec {b}}))^{2}+f_{EW}(|{\vec {a}}-{\vec {b}}|)\times sin(\theta ({\vec {a}},{\vec {b}}))^{2}}$

However, I am unsure if there is a way to sensibly define the angular function ${\displaystyle \theta ({\vec {a}},{\vec {b}})}$. The underlying North-South / East-West functions are suitably constructed such that ${\displaystyle f_{NS}(0)=f_{EW}(0)}$ and ${\displaystyle f_{NS}(d)=f_{EW}(d)=0}$ for all d greater than a limit distance much less than the circumference. Hence I don't have to worry about continuity at ${\displaystyle {\vec {a}}={\vec {b}}}$ or the anitpodes. However, I do also need ${\displaystyle f({\vec {a}},{\vec {b}})=f({\vec {b}},{\vec {a}})}$ which adds an extra constraint to ${\displaystyle \theta }$. I'm not presently sure whether there is a sensible way to construct the angular function that would work over the whole sphere and provide a physically sensible notion of directionality everywhere. Dragons flight ( talk) 01:22, 27 July 2013 (UTC) reply

I'm having difficulty contextualizing the problem. Your functions ${\displaystyle f_{NS}(d)}$ abd ${\displaystyle f_{EW}(d)}$ are not "given" in the normal sense, and that the field depends on the difference of two vectors means that it is not a field on the sphere's surface (i.e. a function of location). You seem to be interested in ${\displaystyle f({\vec {a}},{\vec {b}})}$, where ${\displaystyle {\vec {a}},{\vec {b}}}$ are position vectors of points on its surface. The rest (including ${\displaystyle \theta ({\vec {a}},{\vec {b}})}$) seems to be an attempt at constructing this. Aside from the symmetry requirement, I have no idea what ${\displaystyle f({\vec {a}},{\vec {b}})}$ is supposed to do. Can you build more detail of the picture (without speaking of components) of what it is you're trying to achieve? — Quondum 02:31, 27 July 2013 (UTC) reply

Sorry if I'm being unclear. ${\displaystyle f_{NS}(d)}$ and ${\displaystyle f_{EW}(d)}$ are known, previously specified functions. However, there domains of validity are limited. ${\displaystyle f({\vec {a}},{\vec {b}})=f_{NS}(|{\vec {a}}-{\vec {b}}|)}$ whenever ${\displaystyle {\vec {a}}}$ and ${\displaystyle {\vec {b}}}$ lie on the same line of longitude (i.e. have a North-South orientation). Similarly ${\displaystyle f_{EW}(d)}$ applies only when the points lie on a line of latitude, and hence have an East-West orientation. In the above, the vector norm is meant to indicate the distance on a great circle between the two points. Hence we know a functional form for the special cases of North-South and East-West orientation. We want to define a field that continuously extends this to comparing arbitrary points on a sphere. Such an extension is not unique, and there may well be multiple good choices for how to do this. My expression above in terms of a theta function was meant to be one approach for doing that, though I'm not sure how one could arrive at a natural theta function. Dragons flight ( talk) 03:07, 27 July 2013 (UTC) reply

I need to look at this when my brain is rested, but I will make one observation at this stage: your clarification ${\displaystyle f_{EW}(d)}$ indicates a line of latitude, which is not a great circle except on the equator. This will complicate things somewhat. — Quondum 03:30, 27 July 2013 (UTC) reply

I haven't followed all of the above – but maybe this does what you want. Draw the great circle arc from a to b, look at its bearing at its midpoint, and take its cosine and sine as measures of northness and eastness. Maproom ( talk) 07:15, 27 July 2013 (UTC) reply

Maproom has very neatly found a concept of angle ${\displaystyle \theta }$ that allows the required symmetry (in ${\displaystyle {\vec {a}}}$ and ${\displaystyle {\vec {b}}}$) to be easily achieved. Any combining function must still be symmetric around ${\displaystyle \theta =0}$ and around ${\displaystyle \theta =\pi /2}$, which you have done with the squared cosine and squared sine. This choice of ${\displaystyle \theta }$ is nevertheless still somewhat ad hoc, and will lead to discontinuities when the midpoint of the great arc lies at a pole, thus probably not meeting your smoothness criterion.

In the process of making your formula, you have made a number of rather ad hoc assumptions, which I would call anything but natural, since naturalness (especially in mathematics) usually means exclusion of ad hoc-ness. You have also implied/said that ${\displaystyle f_{NS}({\vec {a}},{\vec {b}})}$ and ${\displaystyle f_{EW}({\vec {a}},{\vec {b}})}$ have limited domains, but by the notation ${\displaystyle f_{NS}(d)}$ and ${\displaystyle f_{EW}(d)}$ you have implied a rather ad hoc extension of these two functions of the points, which you use in combining them. You have also appear to have simply assumed a linear combination of the two functions.

You seem to have a rather ill-posed problem, including by implication that the properties of ${\displaystyle f_{NS}(d)}$ and ${\displaystyle f_{EW}(d)}$ cannot be used: that these could be arbitrary unrelated functions. Their definitions also seem to be rather arbitrarily pinned to lines of longitude and latitude, where a coordinate-free solution might be natural. If ${\displaystyle f_{NS}(d)=f_{EW}(d)}$ (in their respective domains of applicability), then there is a natural extension that is independent of longitude and latitude. I would first try to understand these functions and their properties better before trying too hard for an ad hoc solution. — Quondum 13:55, 27 July 2013 (UTC) reply

How are you defining ${\displaystyle {\vec {a}}}$ and ${\displaystyle {\vec {b}}}$? If ${\displaystyle {\vec {a}}}$ = (a1,a2) := (degrees east, degrees north) etc. then ${\displaystyle \theta }$ = tan^-1((b2-a2)/(b1-a1)) seems to get at the spirit of what you're trying to accomplish, if not the particulars. (I also think that this is more of a math desk question, but since we're here...)-- Wikimedes ( talk) 19:49, 27 July 2013 (UTC) reply

Altering Maproom's idea slightly, the angle a great circle between the two points makes with the equator would be a good measure of the north and east relation of the two points; angles closer to zero indicate a mostly E-W relationship and angles closer to 90 (or -90) would indicate a largely N-S relationship. High N-S relationships might tend to obscure any E-W relationship, say between the point at the intersection of the International Dateline and the Equator (point a) and a point on Greenland (point b), but depending on your application, it might be reasonable to consider Greenland to be mostly north of any point on the equator regardless of longitude. If you want isolated FNS and FEW, you could take the sine and cosine of this angle, respectively. In terms of the sign of the function, if you direct the great circle from point a to point b, and take the angle of (the segment of the great circle immediately following the intersection with the equator) with (the segment of the equator that forms and acute angle with this great circle segment), this should do it. (i.e. the angle from the above mentioned point a to point b would be in the upper right quadrant of the intersection with the equator, but the angle from point b to point a would be in the lower left quadrant and differ by 180 degrees).-- Wikimedes ( talk) 05:40, 28 July 2013 (UTC) reply

I don't follow the way you select between the two equator/great circle intersections, but no matter how you do it, you will have a 180° discontinuity because your selection criterion will switch between them at some point. Also it has the quirk that two points at the same high latitude always produce an angle closer to ±90°, not 0° as you'd expect. — Quondum 21:36, 28 July 2013 (UTC) reply

(ec) That would be because it's poorly defined (yes there are two points of intersection) and I believe my example is self-contradictory whichever choice you use.-- Wikimedes ( talk) 22:05, 28 July 2013 (UTC) reply

Having thought it through and confirmed at Great circle, I see that points directly east-west of each other do in fact lie on a great circle, which can give very strange results for "eastness" using great circles. I'm not sure if this is a fundamental problem with any definition of "eastness", but it's time for me to get back into my depth.-- Wikimedes ( talk) 23:53, 28 July 2013 (UTC) reply