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So I'm having surgery in two months on my eyes. The surgery is to correct a rather villainous Strabismus that has become noticeable lately (both for myself, damn double vision, and those around me). Apparently my doctor is going to basically detach and reattach muscles attached near the front of the eye that control the ability to see inward. I'm curious, what muscle or muscles are those exactly? Sir William Matthew Flinders Petrie | Say Shalom! 01:26, 11 June 2011 (UTC)[reply]
I'm thinking of an object similar to this made of glass or metal, but more elongated, turn it upside-down, and putting holes or tabs on any of the surfaces so that they could be directly fingered or remotely adjusted (think trumpet valves), and using the entire device similar to a saxophone or like a pan flute. What's the timbre? ~AH1(discuss!) 14:45, 11 June 2011 (UTC)[reply]
Hi. What would a klein bottle sound like as a musical instrument, considering that a mouthpiece is attached to the open end, while holes on any part of the surface that can be fingered either directly or using valves adjust its pitch? Thanks. ~AH1(discuss!) 02:16, 11 June 2011 (UTC)[reply]
I don't see how you could use a klein bottle as a horn or similar instrument. There's nowhere for the air to go once you blow into it. I suppose you could use it like a jug though. In which case, it would depend on the interior dimensions of the bottle. More space = deeper tone. Dismas|(talk) 02:40, 11 June 2011 (UTC)[reply]
A Klein bottle is a topological entity. The musical characteristics of a container depend on its resonances, which are essentially independent of its topology. So, the question doesn't really have an answer. (Also it's impossible to embed a genuine Klein bottle in 3D Euclidean space, but that's another issue.) Looie496 (talk) 02:53, 11 June 2011 (UTC)[reply]
You don't have to be able to blow through something to use it as a musical instrument, you can blow across it, as a closed tube, eg a pan flute. Mitch Ames (talk) 03:09, 11 June 2011 (UTC)[reply]
I thought I said that. Dismas|(talk) 04:33, 11 June 2011 (UTC)[reply]
Can someone please direct me to an article that discusses how a space craft is navigated between planets, and between galaxies? I can find a number of Wikipedia entries on space travel, orbital dynamics, etc, but nothing that explains how a space craft can recognise where it is in relation to (say) Earth, what direction it is travelling, and what course corrections it needs to make in order to arrive at where it's destination will be when it gets there, be it another planet or another galaxy.58.174.69.136 (talk) 06:31, 11 June 2011 (UTC)[reply]
Also note that intergalactic travel is really not realistic because of the incredibly large distances. Dauto (talk) 06:59, 11 June 2011 (UTC)[reply]
Dawn sun tracker is the pink rectangle on top, seven others are on each other corner
All space crafts I know of use a Star tracker and a sun tracker. A light sensor looks for the sun which is very easy, for example the Dawn (spacecraft) has a sensor at every corner. The Camera system makes an image of the stars and compares the image with stored maps. With this information it is clear where you are and the where you go is easy to know, because you always go on a straight line (plus gravity) if you do nothing, which is the case most of the time. --Stone (talk) 11:10, 11 June 2011 (UTC)[reply]
How about interstellar travel, or when you lose sight of the sun? Plasmic Physics (talk) 12:57, 11 June 2011 (UTC)[reply]
Just for clarity - no human-built space probe built to date has ever needed a guidance system for interstellar travel. So far, only a couple of our space probes have ever left the vicinity of our Sun; and it's fair to say that they are unguided spacecraft at this time. For example, Voyager II is widely regarded to have transmitted the solar magnetopause and is by some definition now "interstellar" - but the craft has not made an orbital trajectory correction in something like a decade or two. Currently, we here on Earth know where Voyager II is because we track it from Earth using powerful, specialized RADAR (Deep Space Network) - but up there, Voyager II probably doesn't know exactly where it is. It's guidance systems aren't designed for this phase of its lifetime. Its limited computer programs and guidance were intended to control its planetary flyby stages, not post-heliopause navigation. All that the spacecraft is currently aware of are a few readings from a few still-functional scientific instruments. It may still have attitude awareness, but probably has essentially no positional awareness. Nimur (talk) 17:42, 11 June 2011 (UTC)[reply]
Note, everything mentioned so far is for working out what direction the spacecraft is pointing in. You also need to know where you are, but that isn't too difficult. You can just see what direction the radio signals from the spacecraft are coming from and how long they take to get to the Earth (or a relay craft if one is being used). --Tango (talk) 17:36, 11 June 2011 (UTC)[reply]
Your question is entirely hypothetical because an infinite mass does not exist, and an infinite force also does not exist. Science does not concern itself with things that are known not to exist.
It is often claimed that the universe has an infinite amount of mass. Hence, according to these claims, the mass of the universe is infinite.--Leptictidium (mt) 07:43, 11 June 2011 (UTC)[reply]
The universe may have infinite mass, but it's hardly a rigid object. Supposing you had a way of applying infinite force, where exactly would you hook it up to? --Trovatore (talk) 08:04, 11 June 2011 (UTC)[reply]
I was actually thinking of the Big Bang. If the universe indeed had an infinite mass, it would've taken an infinite force to initiate its expansion, wouldn't it?--Leptictidium (mt) 08:10, 11 June 2011 (UTC)[reply]
So I have to admit I don't understand the Big Bang in detail, but I don't think you can apply notions like "force" to it in exactly the same way you do in the current era. Those things get confusing when you start talking about things on the scale of universal expansion. At the time of the Big Bang, everything is on the scale of universal expansion. So for example I have only the vaguest notion of what might have caused inflation (cosmology); maybe there's more information at that article. --Trovatore (talk) 08:47, 11 June 2011 (UTC)[reply]
You say it is often claimed that the universe has an infinite amount of mass but you haven't commented on who makes that claim. It isn't a scientist or a mathematician. The mass of the universe is extremely large - so large that we have no way of assigning a number to it, or knowing exactly how large it is, but this does not make it infinite. Most of the universe is empty space and if the mass of the universe was to be infinite all that empty space, all of it, would have to be filled with mass and both the empty space and the mass would have to be increasing continuously. Clearly that isn't the situation so it is incorrect to apply the mathematical concept of infinite to the universe. Dolphin(t) 08:14, 11 June 2011 (UTC)[reply]
Hmm? No, that's not so. If the universe has infinite volume, then it can have arbitrarily small positive density and still have infinite mass. --Trovatore (talk) 08:36, 11 June 2011 (UTC)[reply]
Thanks for your answer, Dolphin. I actually knew everything you're telling me, but forum members at Physics Forums made me doubt by saying that the volume of the universe is simply more infinite than its mass. From what you're telling me, the forum members' claim is plain gibberish. Leptictidium (mt) 08:20, 11 June 2011 (UTC)[reply]
No, Dolphin is simply wrong. Whether the universe is finite or infinite is not known. --Trovatore (talk) 08:40, 11 June 2011 (UTC)[reply]
I apologize to Dolphin for my tone in the above. I allowed myself to get upset. --Trovatore (talk) 19:37, 11 June 2011 (UTC)[reply]
Physics Forums works by arguments from authority, don't trust a word they are saying on face value. I'm not saying that they are wrong, but they still managed to ban me indefinitely last year, simply because they could not stand that I would very, very occasionally contradict what the "mentors" were saying. What they do is very effective at keeping cranks out, but over time the forum has become similar to Libya under Gaddafi's rule. Count Iblis (talk) 16:51, 11 June 2011 (UTC)[reply]
Hmm, I think that article could use work. For example, we learn from it that [i]f there exists an irresistible force, it follows logically that there cannot be any such thing as an immovable object, and vice versa. But is that really so? Why can't there exist both an irresistible force and an immovable object, but for some reason the force cannot be applied to the object? For example, maybe the force is attached to an unstoppable object that happens to be moving away from the immovable object. The article doesn't seem to be well-sourced; I'm sure someone willing to put in the work could find a more in-depth analysis somewhere. --Trovatore (talk) 09:31, 11 June 2011 (UTC)[reply]
Better sources are always appreciated, but I believe the issue is one of definitions. What do you mean by "immovable object"? Usually you mean "an object that nothing (if it tried) could move". What do you mean by "irresistible force"? Usually "a force which will move anything (if it were applied to it)". So if the "irresistible force" is applied to the "immovable object", and the latter moves, it really wasn't an "immovable object". Conversely, if it *doesn't* move, then it really wasn't an "irresistible force". The only way around that is to alter the definitions (e.g. by dropping the implication of the parenthetical elements). But then you're talking about something different, and would need to be clear what exactly you meant by "irresistible" and "immovable". ("I once knew a dog which danced professional ballet" is surprising until you're told by "a dog", the person means their ex-boyfriend.)
Likewise, the original poster's question hinges on how they come by (the definition of) their infinities. In physics, "what happens when I combine two infinities" is usually referred to a "renormalization", which can yield sensible results, but which depends on where the infinities are from and how they're combined. For example, with a = F/m, we could have and , which would give one answer (no acceleration), or we could have and , which would give another (a finite, non-zero acceleration). -- 174.31.219.218 (talk) 16:03, 11 June 2011 (UTC)[reply]
Slight correction, the mass of the Observable universe is not infinite. The entire universe seems to be much larger and so could easily have an infinite extent, infinite mass and an infinite number of flame wars. Hcobb (talk) 16:59, 11 June 2011 (UTC)[reply]
Statements that we simply don't know the extent of the universe, and we don't know whether it is infinite or not, and therefore we should disregard the fact that the observable universe is not infinite, look to me to be a bit like saying We don't know whether Newton's Laws of Motion hold true at the farthest extremes of the universe and therefore we should not be claiming that Newton's Laws of Motion are true. Dolphin(t) 05:49, 12 June 2011 (UTC)[reply]
The theory of universal expansion predicts that the observable universe must be finite, whether the full universe is finite or infinite. Therefore the finiteness of the observable universe gives us no information about whether the full universe is finite or infinite.
On the other hand, if the curvature of space were reliably measured to be negative (it has not been, thus far; all accepted confidence intervals for the curvature include the value zero), then your very uniformitarian argument would be a point in favor of an infinite universe, for why should the curvature turn positive only far from us, to allow space to wrap back on itself?
I should note that negative curvature does not require an infinite universe; there has been some interesting research into what might be the topology of a compact universe with negative curvature. To my naive eyes, though, these seem rather forced, kind of epicycle-ish. --Trovatore (talk) 07:40, 12 June 2011 (UTC)[reply]
Well, no, we actually do have evidence of the existence of a very large finite mass in the universe, while there is neither any evidence for an infinite mass nor any way even of coherently connecting that idea to any real known observational truth. We simply don't take such ideas seriously, or try to refute them, any more that you would be expected to have to prove that you are not a murderer in response to someone who asserted the mere theoretical possibility of the accusation without bothering to specify who youd had killed or when you had killed them. μηδείς (talk) 21:46, 12 June 2011 (UTC)[reply]
For the first point, there is a very large finite mass for which we have direct evidence; that is not the same as direct evidence that it is finite.
As for "real known observational truth": what is known is subject to change. The error bars on the curvature of space currently include zero. Someday they may be tightened to the point that they do not, and they may lie on the negative side of the line. In that case, the simplest models of spacetime would coherently connect that idea to an infinite universe. Are you suggesting that we should not try to refute that idea by tightening the error bars? --Trovatore (talk) 22:04, 12 June 2011 (UTC)[reply]
So you yourself say that "there is a very large finite mass for which we have direct evidence" yet ask me for "direct evidence that it is finite." That reminds me of an Aristotle quote, something about the fool asking for the proof of the prior by means of the consequent. I will have to post that on the humanities board and get back to you. μηδείς (talk) 02:52, 13 June 2011 (UTC)[reply]
Perhaps my wording was not ideal. The referent of the last it is "all the mass in the universe". That is, we have direct evidence for the existence of a very large finite mass, for example all the mass in the observable universe. We do not have direct evidence, as yet, that the mass in the whole universe is finite.
I can see on rereading that it is possible to interpret the it as referring to the "very large finite mass", and that would indeed give a very strange pair of propositions to assert simultaneously, but that is not what I meant. --Trovatore (talk) 04:35, 13 June 2011 (UTC)[reply]
What nature is, then, and the meaning of the terms 'by nature' and 'according to nature', has been stated. That nature exists, it would be absurd to try to prove; for it is obvious that there are many things of this kind, and to prove what is obvious by what is not is the mark of a man who is unable to distinguish what is self-evident from what is not. (This state of mind is clearly possible. A man blind from birth might reason about colours.) Presumably therefore such persons must be talking about words without any thought to correspond.
Fine, that quote was relevant to what I wrote if you took the referent of the word it to mean the "very large finite mass", which admittedly by the rules of the English language it could be. Normally, however, I tend to count on the fact that, if I say something that, taken completely literally, makes no sense, my interlocutors will spend at least a little effort trying to figure out what I actually meant. --Trovatore (talk) 01:03, 14 June 2011 (UTC)[reply]
I'm studying physics on a very basic level with a book and without a teacher, and now there is a line of reasoning in my book that I don't understand. The book says, summarized:
"A projectile thrown horizontally will in the first second fall a vertical distance of 5 meters below the straight-line path it would have taken without gravity. The curvature of the Earth is such that its surface drops a vertical distance of nearly 5 meters for every 8000 meters tangent to its surface. Thus, a stone thrown fast enough to go a horizontal distance of 8000 meters during the 1 second it takes to fall 5 meters, will orbit Earth. So we see that the orbital speed for close orbit about Earth is 8000 m/s."
Now, I understand how this would work during the first second. But what happens with the acceleration of gravity??? Because during the 2nd second a projectile thrown horizontally will fall 15 meter. How come the projectile keeps orbiting and doesn't crash during the 2nd second??? Lova Falktalk 08:21, 11 June 2011 (UTC)[reply]
I'm too lazy to do the numbers, but from the principle, in the second second, the 8000m line segment has a slight angle with respect to the first. With respect to the second, the first one goes "upward" and the and gravity bends it down for the next segment to be tangential again. Note that in this picture the circular orbit is approximated by straight line segments. To get good results, the approximation would use ever more but shorter segments. 5BYv8cUJ (talk) 09:49, 11 June 2011 (UTC)[reply]
I did a quick sketch (at right, click to zoom) of the situation. The black curve is a segment of a circle and represents the surface of the earth. The gray line is a horizontal tangent at height equal to zero units. What one finds is that the surface of the circle locally approximates a parabolic curve; at time t=1 it has bent away from the horizontal tangent by 1 unit of vertical height; at time t=2 there is a separation of 4 units of height, at t=3 the distance between the tangent and the circle is roughly 9 units. In other words, the rate at which the surface of the earth moves away from the horizontal tangent increases with o distance from the starting point.
What this sketch doesn't account for is the fact that the direction of gravitational acceleration changes as our hypothetical orbiting object changes with time. (Having a constant direction for 'down' is fine when one deals with objects travelling well below orbital speed – a ball tossed in the air really does follow a parabolic path from start to finish – but it doesn't work at all for paths that approach the size of the planet.) From the sketch, it's plain to see that at later time points the gravitational force being exerted won't be straight towards the bottom of the picture. At each point in time, there is a new 'down'; this is what bends the path of the orbiting object into a circle instead of a parabola. TenOfAllTrades(talk) 14:22, 11 June 2011 (UTC)[reply]
(edit conflict) Picture the projectile being far above the Earth's surface, travelling fast enough to maintain a roughly circular orbit, rather than either crashing toward the Earth or flying off into space. At this velocity, the object will experience centripetalacceleration, such that the orbit will follow the curvature of Earth, where every 8 km of forward motion experiences 5 metres of motion toward the Earth's core. Assume also that a hypothetical straight-line velocity, ie. one tangential to a circular orbit with a given radius distance from Earth's core from the object's position, will experience no accelerational "tugging" effect toward the Earth. Conversely, such an object with no forward motion will fall toward the Earth in full gravitational acceleration (9.8 m/s2), given negligeable air resistance. Thus, in low earth orbit, the Earth's acceleration is largely applied unto the object's forward motion, keeping its trajectory the same distance to the Earth's core, which in fact does not accelerate the object, but keeps it at constant velocity. In other words, there are vector components of acceleration (down and forward), but along the projectile's velocity there is no +/- acceleration. Hope this helps. ~AH1(discuss!) 14:37, 11 June 2011 (UTC)[reply]
The key point here is that acceleration is a change in velocity, not just a change in speed. Velocity, and thus by extension acceleration, is a vector, not a scalar, so it has a direction as well as a magnitude. In a constant orbit, even though you're not experiencing a change in speed, you *are* experiencing a change in velocity, as the direction of movement changes. Gravity, acting as a centripetal force, causes an acceleration which rotates the velocity vector, but doesn't change its magnitude. -- 174.31.219.218 (talk) 15:21, 11 June 2011 (UTC)[reply]
Another way to explain this problem: the book is trying to explain a change in height for each second. But, as you intuitively know, the projectile isn't falling all that distance during one single instant - it's falling continously. To really describe the height, and velocity, of the projectile as it flies it's trajectory, we must use a little bit more sophisticated physics - that is, the mathematics must handle the height and velocity as continuous functions of time. Ultimately, this is the the most basic and fundamental application of simple calculus - and it is this problem that forced Isaac Newton to formulate the kinematics of falling objects using calculus. Basically, the book is describing a Riemann sum to approximate height at the end of each second, while in fact height changes continuously during each second. When we treat the function continuously, the math is actually easier to compute (even though a more complex process is being conceptually described). So we simply write that velocity is the first derivative of height, and (here is the most important contribution that Newton made to simple kinematics) ... Acceleration is the second derivative of height ... and acceleration is due to gravity only. You know this when we phrase it as "F = m a" but we are now going to write it as "the height at any time is equal to the integral of velocity, and the velocity is equal to the integral of the acceleration of gravity.". Now, because we're near Earth's surface, gravity is almost constant, so the height is a simple parabola. But in your more sophisticated "orbital cannon" case, the problem becomes more complicated - force due to gravity is changing with position - so we have to compute a more challenging integral! Saving you the complexity of that slightly more difficult Calculus problem, you can accept that the solution to that math problem is a conic section - it can be a parabola, a hyperbola, or an ellipse. If the trajectory intersects with the surface of the planet, the object falls back and hits the Earth. If not, the object becomes orbital, and either stays bound to the Earth or reaches escape velocity and never returns. Our article on orbit illustrates each of those cases. If you want to solve this math for yourself, you can - the simple two-body problem walks you through a standard solution. As you add more complicated effects, like perturbations in the gravity field due to the Moon, Sun, Jupiter, or the imperfect/not-quite-spherical Earth, you must solve a very difficult orbital dynamics math problem, usually with the help of a powerful computer. Nimur (talk) 17:27, 11 June 2011 (UTC)[reply]
I get it now. Thank you all so very much!! Lova Falktalk 17:55, 11 June 2011 (UTC)[reply]
Some unscrupulous food producers in Taiwan have been adding plasticisers to food. Some of the plasticisers include Di-isodecyl phthalate, DIDP; Di-(2-ethylhexyl) phthalate, DEHP; Di-n-octyl phthalate, DNOP; Di-isononyl phthalate, DINP; Di-n-butyl phthalate, DBP; Butyl benzyl phthalate, BBP. The government responds by burning al the contaminated products in incinerators.[1] Is this a good way to destroy the products? Would the plasticisers escape in the air and pollute the environment? F (talk) 09:50, 11 June 2011 (UTC)[reply]
That depends on the temperature. But I think the main purpose is destroying the food, so no one will eat it. I don't think that low quantities of such stuff are dangerous unless you eat it. From the names they look like organic molecules (in the sense of organic chemistry) and will be decomposed by nature (UV-radiation, weathering, micro-organisms), just as the many toxins produced by a large number of poisonous plants. 5BYv8cUJ (talk) 10:37, 11 June 2011 (UTC)[reply]
Generally, temperatures high enough to incinerate the food would also be high enough to destroy most other organic substances, including phthalates. -- An American ultranationalist 67.169.177.176 (talk) 01:25, 12 June 2011 (UTC)[reply]
Plasticizers are organic compounds (in the chemical sense), just like plastics and most food components. They'll burn like all the rest, if subjected to sufficient temperature. The one concern is if they volatilize and escape before combustion. But typical garbage incineration not only handles such compounds regularly, but actually creates a large number of hazardous chemicals during the combustion process. All modern waste incinerators will be fitted with emission control systems and scrubbers which will ensure that what goes up the smokestack contains only a minimal amount fo dangerous compounds, if any. -- 174.31.219.218 (talk) 18:25, 12 June 2011 (UTC)[reply]
Wikipedia says ( Postulates of quantum mechanics) the Ket-Vectors were elements of some Hilbert space, wheras Dirac (The Principles of Quantum Mechanics, 4th edition, page 40) says "The space of bra and ket vectors when the vectors are restricted to be of finite length and to have finite scalar products is called by mathematicians a Hilbert space. The bra and ket vectors that we now use form a more general space than a Hilbert space."
To resolve that contradiction I assume that some later development has put all of Diracs intuitive integration and the need for infinity as a number (with all its ensuing troubles and inconsistencies) into a solidly defined mathematical framework that turned out to really be a proper Hilbert space. The question is now: where can I find this solid definition? 5BYv8cUJ (talk) 10:26, 11 June 2011 (UTC)[reply]
Not infinity as a number per se, but the theory of distributions had yet to be set up. One could then use that re-define the Hilbert space you need in quantum mechanics when you want to work in an infinite volume. You then get the Rigged Hilbert space. Note that you don't actually need to know about this to do computations in quantum mechanics. Count Iblis (talk) 15:12, 11 June 2011 (UTC)[reply]
I remember a talk long ago when someone stated that students of mathematics can't do computation. For example, they cannot compute the inverse of a matrix, whereas students of physics can. Someone answered, yes, that's true. But, on the other hand, students of physics can compute the inverse for any matrix.
Having told this anecdote I hope you understand what I have in mind if the following looks like nitpicking or trolling. Dirac thinks he needs vector lengths and scalar products to be allowed to be infinite, and vector lengths and scalar products should result to give numbers (elements of a commutative field), because otherwise all the theorems for vector spaces (and basic computation, too) would have to be prooven again. Every try to do computations with infinity leads to a hell of trouble, the least of which is me proving that zero equals one, using only the contradicting assumptions. Now your link to the rigged Hilbert space doesn't give me the mental breakthrough. Just for any one simple example from quantum mechanics, how exactly would that H and that really be defined? 5BYv8cUJ (talk) 16:15, 11 June 2011 (UTC)[reply]
While I have studied functional analysis, I don't know much about rigged Hilbet spaces. It is actually pretty useless in theoretical physics. You can avoid using this and still be rigorous, by putting systems in finite volumes (take the limit of V to infinity at the end of computations), avoid working in the ill-defined position basis (even if you can define it using rigged Hilbert spaces, physically it is not well-defined). Count Iblis (talk) 15:18, 12 June 2011 (UTC)[reply]
I have been rereading the book (up to page 40 where I throw a "NotANumberException"). I thought of a Schwartz space to fix it, which is not exactly what you are proposing but runs in the same lines. But I don't see how or even if either of those would cure the problem. For what I have seen, the only reason those infinity-values are needed is to fix problems that arise when he does what he calls "superposition" in the introduction and needs to do it on more than countably many kets, which in turn he believes to be necessary to represent any ket from the "complete sets" of eigenvectors of an observable. I am sure modern maths can give those intentions a solid base somehow, but I get crazy because I can't find such a thing anywhere. 5BYv8cUJ (talk) 15:51, 12 June 2011 (UTC)[reply]
You should look at the proof of completeness of bases of eigenfunctions in functional analysis textbooks. The case of discrete spectrum is rather simple, the case of the continuous spectrum is more complicated. But then, if you limit yourself to systems in a finite volume, you only have to deal with the discrete case. Also, physically, an infinite volume is in practice always a finite volume which is very large, so this is understood as the limit of V to infinity. Count Iblis (talk) 16:08, 12 June 2011 (UTC)[reply]
I have recovered that book yesterday and was shocked how much I have forgotten. I'll have a look at spectral theory and see if this gives me the hint I'm lacking now. 5BYv8cUJ (talk) 16:39, 12 June 2011 (UTC)[reply]
Did the early airline aircraft (e.g. 1920s) have toilets/lavatories, at least on longer flights?--68.175.35.188 (talk) 18:45, 11 June 2011 (UTC)[reply]
I am in orbit round a planet and am a few hundred meters behind my command module at the same altitude. I need to dock with the command module. What combinstion of forward and reverse thrust should I use? (Not home work-- just interested).--78.150.233.171 (talk) 19:45, 11 June 2011 (UTC)[reply]
To quote Larry Niven, "Forward takes you out, out takes you back, back takes you in, and in takes you forward."[2] I don't know how literally true this is, but it seems to make sense... ;) Wnt (talk) 22:25, 11 June 2011 (UTC)[reply]
If you think about it, it would have to be a combination of first thrust TA (say) in direction DA (say) to introduce relative movement that decreases the distance between your module and the command module, followed by no thrust as the relative velocity between continues to decrease the distance, followed by reverse thrust TB (say) in directon DB (say) to bring the relative velocity back to zero. Hopefully exactly at the time the two are docking together. I would suspect that thrust TA = TB, and direction DA = - DB. I can't prove it though.--InverseSubstance (talk) 23:37, 11 June 2011 (UTC)[reply]
InverseSubstance is correct -- you need to use forward thrust to speed up and close the distance, followed by an equal amount of reverse thrust to bring the speed back down to the same level as before. You also need to apply a small amount of downward thrust (toward the planet) at the same time as the forward thrust, and upward thrust at the same time as the reverse thrust, in order to compensate for the small change in orbital radius that is caused by the change in speed. Or alternatively, you could use downward thrust alone to speed up and the upward thrust alone to slow down, which is essentially equivalent to doing a low yo-yo in a fighter plane to catch up to your target without overshooting it (though the latter admittedly looks a lot more dramatic). 67.169.177.176 (talk) 01:47, 12 June 2011 (UTC)[reply]
Using forward thrust won't work. It has the paradoxical effect of slowing you down because it will bring you to a higher orbit and for every unit of energy provided by the thrust, two units of energy get "stolen" by gravitational potential energy effectively removing one unit of energy from your kinetic energy and slowing you down. What you need to to is break! That will get the low yo-yo maneuver you described started. Dauto (talk) 05:33, 12 June 2011 (UTC)[reply]
"Flying a spacecraft is very different than flying a plane. There is no true up or down and the dynamics of orbital flight make maneuvering to dock, or rendezvous, two spaceships very complex. I focused my research on solving the problems of speed and centrifugal energy which lead to an ‘orbital paradox’ – a situation in which a pilot who speeds up to catch another craft in a higher orbit will end up in an even higher orbit, traveling at a slower speed and watching the second craft fly off into the distance. The solution to this paradox is counter intuitive, and required new orbital mechanics and procedures. Later, after joining the NASA astronaut corps, I spent time translating complex orbital mechanics into relatively simple flight plans for my colleagues – they thanked me (with a mixture of respect and sarcasm) with the nickname Dr. Rendezvous." Buzz Aldrin.
That is a very interesting aspect. What if the acceleration is done by gravitation? OK, you might not want to wait for this in a space ship, but I'm thinking about Theia_(planet). 5BYv8cUJ (talk) 06:54, 12 June 2011 (UTC)[reply]
Those kind of collisions happen by random chance. If you have hundreds of planetoids in similar orbits then, over millions of years, some of them are inevitably going to intersect. --Tango (talk) 13:00, 12 June 2011 (UTC)[reply]
I was thinking about if it was predictable that Theia hit earth from a slightly increased orbit thus accelerating earth's rotation instead of de-accelerating or reversing it. 5BYv8cUJ (talk) 16:33, 12 June 2011 (UTC)[reply]
Yes, Theia must have increased earth's rotation otherwise the moon wouldn't have a prograde motion. Dauto (talk) 03:32, 13 June 2011 (UTC)[reply]
(un-indent) The space shuttle faces this exact same problem when docking with the ISS. For the final part of the shuttle docking profile the shuttle starts a couple hundred meters ahead of the ISS at the same altitude, and the same velocity. To move towards the station the pilot maneuvers the shuttle radial out, that is away from the Earth. This temporarily moves the shuttle to a higher altitude where it has a slower velocity than the ISS, thus closing the distance between the two. The effect is that the final approach profile is a series of hops, as you can kind of see in this photo. In your situation the active vehicle is behind the passive vehicle, so you will need to move radial-in (towards the Earth), dropping into a lower, faster orbit to close the distance with the passive vehicle. anonymous6494 14:37, 13 June 2011 (UTC)[reply]
Yes, thank you Dauto and Anonymous, you are correct. I was thinking more as an airplane pilot (which I am) rather than an astronaut (which I'm not) when I first answered the question. So my first answer (forward thrust) was wrong, while the second one (the "low yo-yo") was indeed correct. I also assumed, from the way the OP phrased the question, that the spacecraft has provision for thrust straight up and straight down (I meant radial out and radial in -- see what I mean when I say that I can't stop thinking like an airplane pilot?) as well as forward/reverse. 67.169.177.176 (talk) 21:42, 13 June 2011 (UTC)[reply]
Is the Pythagorean scale universally pleasing?edit
Pythagoras said that musical strings of lengths in simple ratios (1 through 4) will produce "pleasing" sounds. Are only people in Western cultures (beginning with the Greeks) pleased by such sounds, or do all people people find the sounds pleasing? 82.31.133.165 (talk) 20:29, 11 June 2011 (UTC)[reply]
I don't know about what people find pleasing, but Musical scale#Non-Western scales will tell you a bit about what different cultures use in their music. It sounds like most do use harmonic scales (which is Pythagoras was talking about), but apparently Indonesia doesn't. It doesn't go into any detail, though. --Tango (talk) 23:59, 11 June 2011 (UTC)[reply]
From personal experience I would say the human ear / brain combination, adopts with time to new scales, as with languages. Newly heard music in an previously unknown scale or system may sound strange at the beginning, but eventually one finds it very pleasing. Take for example the music of Arnold Schoenberg or Free Jazz which are both in western scales, but may sound strange at first. Or Persian Music or Arabic Music or Greek Music which use different scales. Also note that the "western" scales (equal temperament) that we use today are quite different from the scales used a few hundred years ago, like the well temperament, kirnberger or werckmeister temperament. --helohe(talk) 00:07, 12 June 2011 (UTC)[reply]
Also note that not everyone likes Schoenberg or free jazz -- in fact, there are a lot of people who absolutely hate it. This has to do with personal and cultural preferences more than anything else. 67.169.177.176 (talk) 01:35, 12 June 2011 (UTC)[reply]
I'm afraid that includes me! Doesn't matter how long I listen. Alansplodge (talk) 18:02, 12 June 2011 (UTC)[reply]
Note that Schoenberg and free jazz (and other modern western composers inserting smaller dissonaces in their work) are intentionally trying to be jarring to their listeners by not using the "well-tempered" notes that everybody used in earlier works. 75.41.110.200 (talk) 14:33, 12 June 2011 (UTC)[reply]
Our inferior colliculus is hard-wired for the processing of low-order harmonics, which is why harmonic tone combinations are widely preferred in music.[3]Red Act (talk) 01:47, 12 June 2011 (UTC)[reply]
Simple pleasant harmonies not only appeal to humans but to most animals as proof of health. Dislike of dissonances compared to consonance has nothing to do with learning. Human preference for simple harmony, especially the octave, is universal. Helmholtz. μηδείς (talk) 22:46, 12 June 2011 (UTC)[reply]
virtual particle exchange for attraction forcesedit
How does the virtual particle exchange work for attraction? For example electromagnetic attraction between a proton and an electron. I mean, the virtual particle would need to have negative mass, or be emitted in the opposite direction (away from the centre of both particles). I couldn't find a good explanation anywhere. --helohe(talk) 23:09, 11 June 2011 (UTC)[reply]
Are you talking about gauge boson theory ? Yes, it's extremely non-intuitive, but then so is most of quantum dynamics. Even worse than what you mentioned is that a nearly infinite number of virtual particles would be required, especially in the case of gravitons, where a continuous stream of gravitons is needed from every atom to every other in the universe. That's a lot of gravitons. Myself, I think it might be a useful model, in some cases, but don't think it's literally correct (something like how the Dalton model of the atom was useful, but later needed many refinements). StuRat (talk) 23:46, 11 June 2011 (UTC)[reply]
Yes I meant gauge boson theory. I'm also wondering how the virtual particles "detect" the presence and the position of the involved particles. I would find it much more intuitive to understand if all virtual particles would be replaced with fields (or space-time "curvature" like in general relativity). --helohe(talk) 00:00, 12 June 2011 (UTC)[reply]
I think the idea is that there's so darned many virtual particles that those given off by each particle are bound to hit every other particle in the universe. Yes, this seems silly to me, too, and a field or wave does seem like a better approach. However, due to wave-particle duality, perhaps the bosons are sometimes particles and sometimes waves. StuRat (talk) 00:08, 12 June 2011 (UTC)[reply]
Virtual particles can have negative mass. Physicists refer to that by stating that virtual particles are not constrained to their mass shell. But note that any attempt to understand QM in terms of classical concepts is bound to fail at some point. Dauto (talk) 00:09, 12 June 2011 (UTC)[reply]
Compare with tunneling. In quantum mechanics an intitial state can evolve to some final state that classically would be impossible, because the system would have to evolve via an intermediary state which would have an inconsistent energy and momentum. In this case, the charged particles couple to photons, and the "forbidden" intermediary state is the virtual photon, which thus carries energy and momentum in a way that is unphysical. Count Iblis (talk) 15:29, 12 June 2011 (UTC)[reply]
The momentum is off-shell and unphysical whether the force is attractive or repulsive, so I don't think that was the question. Also, the off-shell momentum transfer is the same in classical electromagnetism, and loop-free Feynman diagrams can be used to describe classical electromagnetism, so I don't think there's anything essentially quantum about the question. -- BenRG (talk) 18:53, 12 June 2011 (UTC)[reply]
Not a spacetime diagram.Quantum field theory is a theory of fields, as the name suggests. Virtual particles show up in Feynman diagrams, which are a useful way of doing some calculations in quantum field theory, but don't work at all for others. I don't think it's very helpful to say that fields (like the electromagnetic field) are caused by virtual particles, because, for one thing, that picture doesn't even work in general.
Feynman diagrams look like spacetime diagrams, but they aren't. They are actually just graphs, that is, vertices connected by edges. The position of the vertices, and the orientation and length of the edges, don't mean anything. In a spacetime diagram, the momentum of a particle is always tangent/parallel to its worldline, but in a Feynman diagram you can assign any momentum to any edge as long as momentum is conserved at each vertex. So attractive forces don't present any difficulty. It's hard to explain why like charges repel and unlike charges attract, but it's easy to explain how Feynman diagrams can accommodate the attractive case: it's the same as the repulsive case because directions are meaningless. -- BenRG (talk) 18:53, 12 June 2011 (UTC)[reply]
I agree with every thing you said except that Feynman diagrams can be seen as abstract space-time diagrams. For instance, in that penguin picture to the right time is usually taken to flow to the right, in the sense that the lines on the left represent incoming particles while the lines on the right represent out-going particles, even though the intermediary vertices may happen at any point in time or space (even if that requires fast than light travel). Dauto (talk) 22:19, 12 June 2011 (UTC)[reply]
I think you are asking if it takes more energy to get a temperature change started than to keep it going in the same direction. In that case, the answer is "no, the first degree of temperature change takes about the same energy as the next". An exception would be if the first degree requires a phase change. StuRat (talk) 23:44, 11 June 2011 (UTC)[reply]
What about the time required to initiate the heating, assuming constant energy input into the fluid? Or is any "inertial" effect caused by the starting time required of the heating device itself? ~AH1(discuss!) 23:50, 11 June 2011 (UTC)[reply]
Yes, the heating time of the device itself, or whatever is in-between. In the case of a pot on an electric stove, for instance, first the burner must heat up, then it must radiate that heat to the bottom of the pot, then that heat must conduct to the inside of the pot. Only then will the liquid in the pot start to heat up. With a thick, cast-iron pot, that could take a while. With a thin, copper pot on a high flame, the heat should make it to the liquid quite quickly. A glass of liquid in a microwave should start to heat essentially instantaneously. StuRat (talk) 23:58, 11 June 2011 (UTC)[reply]
In the case of the microwave, this also has to do with the fact that the liquid is being heated directly by absorbing electromagnetic radiation, rather than by conduction. So essentially, you can think of it as the case where there's nothing in between at all. 67.169.177.176 (talk) 01:30, 12 June 2011 (UTC)[reply]
