 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
Natural Science Forum / Physics / Particle Physics / June 2005New theory shows Fermions have odd spin, Bosons have even spin!
We have all noticed that spin is described as being a multiple of hbar/2. I thought that it would be better to set this value to a constant giving, hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s But now the equations for spin did not work with hdot, so I had to correct them. Here are the corrected equations, |sv| = sqrt(s(s + 2)) * hdot and Sz = ms.hdot where, sv is the quantized spin vector, |sv| is the norm of the spin vector, s is the spin quantum number, which can be any non negative integer, Sz is the spin z projection, ms is the secondary spin quantum number, ranging from -s to +s in steps of two integers For spin 1 particles this give: |sv| = sqrt(3).hdot and Sz = -hdot, +hdot For spin 2 particles this give: |sv| = sqrt(8).hdot and Sz = -hdot, 0, +hdot Now that the spin equations have been corrected, the definitions for fermions and bosons are incorrect, and must be redefined as follows. Fermions are particles that that have an odd integer spin. Bosons are particles that have an even integer spin. Would these redefinitions have any other effects on the Standard Model? Can these redefinitions explain any currently unexplained phenomena? Are there any experiments that could confirm or refute these claims? I would like eveyone to have a good think about this, and give me your objections to it, or even data to support it. Happy shooting :) It may also be of interest to note that the only even prime number is 2. > We have all noticed that spin is described as being a multiple of > hbar/2. I thought that it would be better to set this value to a > constant giving, > > hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s *Plonk* Remember this, hbar is the quantum of angular momentum spin is not angular momentum, so hbar is not the quantum of spin. > Remember this, > > hbar is the quantum of angular momentum You still don't know what that actually means, right? > spin is not angular momentum, Err, it is. Spin is *intrinsic* angular momentum. [snip] Bye, Bjoern > > Remember this, > > > > hbar is the quantum of angular momentum > > You still don't know what that actually means, right? I have answered this before, but will do so again. A quantum is a discrete indivisible value, the smallest value something can have. For example, if something has a quantum of 1, it will only be allowed to have a value which is a multiple of 1. It can not have a value of 0.5 or 1.5. If you disagree with this, then tell me what you think it means. > > spin is not angular momentum, > > Err, it is. Spin is *intrinsic* angular momentum. >From wiki, "Whereas classical angular momentum arises from the rotation of an extended object, spin is not associated with any rotating internal masses, but is intrinsic to the particle itself." Therefore, 'intrinsic angular momentum' and 'angular momentum' are not the same. > [snip] > > Bye, > Bjoern > > > Remember this, > > > [quoted text clipped - 8 lines] > to have a value which is a multiple of 1. It can not have a value of > 0.5 or 1.5. That is incorrect. What is more important in this connection is that it comes in STEPS of 1. I can understand why you think it's natural for something to have spin quantum numbers of 0, 1, 2, especially since the *projection* of the spin is also quantized. So for spin s = 1, the spin projection ms runs from -s to s, or in this case, -1, 0, 1. This is experimentally distinguished by circumstances that energetically separate the spin projection states. If it exhibits three distinct states, we know it is spin 1. But it's also natural for something to have a spin s = 1/2. In this case, ms can have values -1/2 and 1/2, separated by a step of 1, you see. Similarly, the experimental signature is 2 energetically separated spin projection states. If there are two states, then we know it is spin 1/2. With regard to this, look up Stern-Gerlach. > If you disagree with this, then tell me what you think it means. > [quoted text clipped - 9 lines] > Therefore, 'intrinsic angular momentum' and 'angular momentum' are not > the same. No. Therefore, "intrinsic angular momentum" and "classical angular momentum" are not the same. (Reread Wiki.) Only the latter is associated with rotation. Both are associated with angular momentum. PD >No. Therefore, "intrinsic angular momentum" and "classical angular >momentum" are not the same. (Reread Wiki.) Only the latter is >associated with rotation. Both are associated with angular momentum. PD, why are you damn anal retentive? Are you autistic or something? Doesn't angular momentum normally imply rotation? The man is trying to say that intrinsic angular momentum (spin) is not associated with rotation. That's all. You nitpick things to death, man. It's counterproductive. Louis Savain The Silver Bullet: Why Software Is Bad and What We Can Do to Fix it http://users.adelphia.net/~lilavois/Cosas/Reliability.htm > > > > Remember this, > > > > [quoted text clipped - 11 lines] > That is incorrect. What is more important in this connection is that it > comes in STEPS of 1. What part is incorrect? In the above example, I am not talking abot spin, but the quantum of something in general. > I can understand why you think it's natural for something to have spin > quantum numbers of 0, 1, 2, especially since the *projection* of the [quoted text clipped - 3 lines] > projection states. If it exhibits three distinct states, we know it is > spin 1. I disagree, my calculations show that it is spin 2. > But it's also natural for something to have a spin s = 1/2. In this > case, ms can have values -1/2 and 1/2, separated by a step of 1, you > see. Similarly, the experimental signature is 2 energetically separated > spin projection states. If there are two states, then we know it is > spin 1/2. With regard to this, look up Stern-Gerlach. Why is it natural for something to have spin 1/2? A natutal number is any non negative integer, 1/2 is not natural. My calculations show that if there are 2 states, it is spin 1, which is natural. > > If you disagree with this, then tell me what you think it means. > > [quoted text clipped - 13 lines] > momentum" are not the same. (Reread Wiki.) Only the latter is > associated with rotation. Both are associated with angular momentum. In one has different properties than the other, they are not the same. > > > > > Remember this, > > > > > [quoted text clipped - 25 lines] > > I disagree, my calculations show that it is spin 2. Your calculations show nothing of the sort. What you said is: IF the expression for the expectation value for the total spin is changed, and IF it assumed that quantum numbers can *only* be non-negative integers, THEN in this scheme, the spin of a boson is 2. At best, this is a relabeling scheme. > > But it's also natural for something to have a spin s = 1/2. In this > > case, ms can have values -1/2 and 1/2, separated by a step of 1, you [quoted text clipped - 4 lines] > Why is it natural for something to have spin 1/2? > A natutal number is any non negative integer, 1/2 is not natural. And who says a quantum number has to be "natural", according to your feeling of what natural is? > My calculations show that if there are 2 states, it is spin 1, which is > natural. This is based on numerology, not physics. The *physics* calculation says -- after much work, not after pondering what would be "natural" -- that the expectation value of the spin projection, ms, takes on the values -s, -s+1, ... s-1, s. So if s=2, ms takes on possible values -2, -1, 0, 1, 2 (5 possible spin projection states). And if s = 3/2, ms takes on possible values -3/2, -1/2, 1/2, 3/2 (4 possible spin projection states). How many states are there are EASILY experimentally verifiable in a spectrometer. See the spectrometry chapters in any organic chemistry textbook to see an excellent application of this principle (in fact, 60 pages or so of application after application after successful application). > > > If you disagree with this, then tell me what you think it means. > > > [quoted text clipped - 15 lines] > > In one has different properties than the other, they are not the same. That depends on the properties that you select to define the object. A male cardinal has brighter red coloring and a distinct feathered crest, and a female cardinal does not. They are both cardinals, though they have different properties. Angular momentum is defined according to the conservation law it obeys, and how it transforms under a coordinate transformation, not whether it is associated with the rotation of a spatially extensive object. Classical angular momentum often *does* involve a rotation but it is not central to the definition. By the way, did you know that an object that is traveling in an absolutely straight line at constant velocity has *classical* angular momentum with respect to an axis off that line? Where is the rotation there? PD > Your calculations show nothing of the sort. What you said is: > IF the expression for the expectation value for the total spin is > changed, and > IF it assumed that quantum numbers can *only* be non-negative integers, > THEN in this scheme, the spin of a boson is 2. > At best, this is a relabeling scheme. That is what I have been saying. > > > But it's also natural for something to have a spin s = 1/2. In this > > > case, ms can have values -1/2 and 1/2, separated by a step of 1, you [quoted text clipped - 7 lines] > And who says a quantum number has to be "natural", according to your > feeling of what natural is? I don't know. Consider this though, when calculating electron orbitals, n,l and m are all integers, why should s be any different? My method fits in with the existing data perfectly. > > My calculations show that if there are 2 states, it is spin 1, which is > > natural. [quoted text clipped - 10 lines] > principle (in fact, 60 pages or so of application after application > after successful application). No, I i based this on the fact that spin can only be multiples of hbar/2. If a 50 year old person is asked his age, does he say that he is 100/2? Of course not! This is why I created a new constant and called it hdot. hbar/2 is nonsense. I don't know what your'e getting at here. You get the same number of spin projection states from either equation. I don't know much about spectrometry, so I can't comment on this at the moment. > > > > If you disagree with this, then tell me what you think it means. > > > > [quoted text clipped - 20 lines] > crest, and a female cardinal does not. They are both cardinals, though > they have different properties. They may both be cardinals, but are they the same cardinal? Definitely not! > Angular momentum is defined according to the conservation law it obeys, > and how it transforms under a coordinate transformation, not whether it [quoted text clipped - 4 lines] > has *classical* angular momentum with respect to an axis off that line? > Where is the rotation there? Well I would say that the conservation of angular momentum was defined according to observations of angular momentum. Look, angular momentum is a vector, spin is a spinor. Why can you not see that they are diferent? > > Your calculations show nothing of the sort. What you said is: > > IF the expression for the expectation value for the total spin is [quoted text clipped - 21 lines] > all integers, why should s be any different? My method fits in with the > existing data perfectly. I see. So you just want the parameterization changed, multiplying a quantum number by two and changing a constant to absorb the factor of two, because you think it looks better. Then, instead of fermions having half-integer spins, they now have odd-number spin quantum numbers, and instead of bosons having integer spins, they now have even-number spin quantum numbers. Fine. But your subject line then sounds like a National Enquirer headline: "Pamela Anderson involved in Animal Abuse Shocker!" where the story actually says "Pamela Anderson volunteers weekly at the local animal shelter. In an exclusive interview, she told us how sick and thin some animals are when they are picked up from the streets. 'I'm shocked,' Pamela told us." Your subject line should have been more like: "New theory shuffles things around so that fermions have odd spin and bosons have even spin - no big deal, but it's prettier to me." > > > My calculations show that if there are 2 states, it is spin 1, which is > > > natural. [quoted text clipped - 22 lines] > I don't know much about spectrometry, so I can't comment on this at the > moment. That's why experiment and the theoretical underpinnings are more important than numerology. > > > > > If you disagree with this, then tell me what you think it means. > > > > > [quoted text clipped - 23 lines] > They may both be cardinals, but are they the same cardinal? Definitely > not! That of course is irrelevant. Two things do not have to be the very same object to be classified as the same kind of thing. Here it's important to understand what makes a class of things a species. > > Angular momentum is defined according to the conservation law it obeys, > > and how it transforms under a coordinate transformation, not whether it [quoted text clipped - 10 lines] > Look, angular momentum is a vector, spin is a spinor. Why can you not > see that they are diferent? And that's where you're wrong. Classical angular momentum and spin are both pseudovectors (sometimes called spinors). That's the part about "how it transforms under a coordinate transformation" that I was referring to. Sorry, but there really is good, sound reason for calling them the same kind of thing. PD > > > Your calculations show nothing of the sort. What you said is: > > > IF the expression for the expectation value for the total spin is [quoted text clipped - 40 lines] > "New theory shuffles things around so that fermions have odd spin and > bosons have even spin - no big deal, but it's prettier to me." That would be too long to fit on screen. > > > > My calculations show that if there are 2 states, it is spin 1, which is > > > > natural. [quoted text clipped - 25 lines] > That's why experiment and the theoretical underpinnings are more > important than numerology. Is there any experiment that can prove that fermions have half-intger spin and bosons integer spin, as opposed to fermions having odd integer spin and bosons even integer spin? What are the theoretical underpinnngs you are talking about here? Where does numerology come into it? > > > > > > If you disagree with this, then tell me what you think it means. > > > > > > [quoted text clipped - 27 lines] > same object to be classified as the same kind of thing. Here it's > important to understand what makes a class of things a species. They may both be a type of angular momentum, but they behave differently. If they behave differently, then they are differnet, for me it's as simple as that. > > > Angular momentum is defined according to the conservation law it obeys, > > > and how it transforms under a coordinate transformation, not whether it [quoted text clipped - 18 lines] > > PD I just looked up pseudovector, and yes angular momentum is listed as an example, so you got me there. As far as I know, if an object has angular momentum then it is rotating around a point, while spin is not associated with any rotation. Is this correct? Could you tell me what you think is the difference between angular momentum and spin? [snip] > > Your subject line should have been more like: > > "New theory shuffles things around so that fermions have odd spin and > > bosons have even spin - no big deal, but it's prettier to me." > > That would be too long to fit on screen. Would have prevented a tirade, though. > > > > > My calculations show that if there are 2 states, it is spin 1, which is > > > > > natural. [quoted text clipped - 29 lines] > spin and bosons integer spin, as opposed to fermions having odd integer > spin and bosons even integer spin? You miss the point. Physics isn't about whether fermions have half-integer spin or odd integer spin (with spin defined a little differently). Physics is about whether, using a theory with quantized spin, you successfully predict experimental behavior that you could not predict with a different theory. (By "different" I mean one that's not algebraically equivalent.) As you admit, your rescaled version of spin changes nothing about the experimentally observed predictions, it just makes the parameters a little more to your liking. Theories don't attempt to predict their own parameters, they predict observable effects. Indeed, reparametrizing a theory is exceedingly uninteresting, which is why I dismissed it cavalierly. > What are the theoretical underpinnngs you are talking about here? Ah, time to pick up a book. I suggest you start with Eisberg and Resnick. > Where does numerology come into it? When you fiddle with a theory to make the numbers in it cleaner, that's numerology. That's obsessing about the numbers and not about the physics. Look, it's a common convention in particle physics to choose a nonstandard system of units so that hbar = c = 1. This is done *purely* so that all those pesky factors of hbar and c in complicated formulas don't have to be written down. Does that make a new, more elegant theory, because the formulas look simpler? Not on your life. That is decidedly NOT interesting physics, though some students make more of it than it is. > > That of course is irrelevant. Two things do not have to be the very > > same object to be classified as the same kind of thing. Here it's [quoted text clipped - 4 lines] > If they behave differently, then they are differnet, for me it's as > simple as that. That depends on what you classify as characteristic behavior. Some humans have black skin, some have white skin, and the musculature in their calves is decidedly different. Those features are simply irrelevant to the property of their being human. See below. > > > > Angular momentum is defined according to the conservation law it obeys, > > > > and how it transforms under a coordinate transformation, not whether it [quoted text clipped - 25 lines] > around a point, while spin is not associated with any rotation. Is this > correct? No, and this is what I was getting at. If an object has angular momentum, then it has a property that behaves like a pseudoscalar which, in any closed system of such objects, combines linearly to make a total quantity that is conserved no matter what happens internally to the system. That's what makes it angular momentum. Going around in a little circle around an axis is NOT a defining feature of angular momentum. It so *happens* that objects that go around in a little circle have angular momentum, but that's not a defining feature. Consider the thing I mentioned before: things going in a straight line. A plane flying in a straight line by a control tower has angular momentum with respect to the center of the tower; it does *not* have to fly in a circle around the tower for that to be true. If you don't believe that, then let me guide your thinking just a little more. a) The earth going around the sun in nearly a circular orbit has angular momentum about the sun, does it not? b) What about Neptune, that orbits the sun in an eccentric, oval orbit? It has angular momentum about the sun, too, does it not? c) And what about Halley's Comet, which swings in a *highly* eccentric elliptical orbit around the sun? It has angular momentum around the sun, too, doesn't it? d) And what about one-time comets that are drawn into our solar system from very far away, swing around the sun and are slingshotted away on a hyperbolic path around the sun to never return again? Those are no different than Halley's comet except by degree. Those have angular momentum, do they not? e) And what about distant comets whose paths are barely bent by the sun, approaching and then receding from the solar system on almost a straight line. Does that still have angular momentum? I think you can see where I'm going with this asymptotic process. If not all these have angular momentum, then at what point in the continuum is it lost? All this is to point out that even classically, having angular momentum does not mean going around in a circle around an axis. > Could you tell me what you think is the difference between angular > momentum and spin? Spin is a kind of angular momentum. Did you mean what I think is the difference between *classical* angular momentum and spin? PD > [snip] > > > [quoted text clipped - 5 lines] > > Would have prevented a tirade, though. But it's good to talk. > > > > > > My calculations show that if there are 2 states, it is spin 1, which is > > > > > > natural. [quoted text clipped - 119 lines] > does *not* have to fly in a circle around the tower for that to be > true. Yes, but the plane is flying around the earth. > If you don't believe that, then let me guide your thinking just a > little more. [quoted text clipped - 18 lines] > having angular momentum does not mean going around in a circle around > an axis. I never mentioned circle, I said rotating around a point. In all the cases you listed above, it is possible to find the point that these objects are rotating around. > > Could you tell me what you think is the difference between angular > > momentum and spin? > > Spin is a kind of angular momentum. Did you mean what I think is the > difference between *classical* angular momentum and spin? Yes this is what I meant. [snip] > > Going around in a little circle around an axis is NOT a defining > > feature of angular momentum. It so *happens* that objects that go [quoted text clipped - 33 lines] > cases you listed above, it is possible to find the point that these > objects are rotating around. Really? What constitutes rotation about a point? Does it have to complete an orbit? If not, then is a 90 degree deflection enough to constitute rotation? If 90 degrees is enough, is 10 degrees? Is 2 degrees? Is 0.00002 degrees? In the comet example I gave above, cases (d) and (e) probe that limit. Indeed, if I have a hockey puck sliding on the ice and as it goes by I blow on it so that its path is deflected by a degree or so from its former path, is that deflection a rotation about some point? If so, which point? If not, how can you tell the difference between this and the comet that is deflected by 0.05 degrees as it "rotates" around the edge of the solar system, its path perturbed ever so slightly by the sun? What I claim further is that the asymptotic limit, that is, absolutely straight line motion ALSO has angular momentum about an axis off that path. How can I guarantee that it will be straight? Constrain the object to glide on a rail -- I don't care. But in this case, the path is *straight* and there is no center to find. > > > Could you tell me what you think is the difference between angular > > > momentum and spin? [quoted text clipped - 3 lines] > > > Yes this is what I meant. Classical angular momentum associates some mass in tangential motion at some distance with respect to a chosen axis. At least, that's what distinguishes it from spin, just like black fur would distinguish some cats from some other cats. There is more I can tell you about angular momentum, much more. But first think about the things I posed to you above. PD > [snip] > > > [quoted text clipped - 47 lines] > as it "rotates" around the edge of the solar system, its path perturbed > ever so slightly by the sun? In the case of the comet, the comet will complete orbits around the point, unless it is affected by the gravity of some other object, which will then alter that point which the comet is rotating around. >From wiki, "Deflection is the physical event where an object collides and bounces against a plane surface." This is not rotation at all. I don't know what you mean here, but hbar is the quantum of angular momentum if thats what you're getting at. For me, the whole universe is rotating around its center point, so everything in it is also rotating around a point. If this is not the case, then galaxies are rotating around a point, and so, everything in a galaxy is rotating around a point. Everything has angular momentum! > What I claim further is that the asymptotic limit, that is, absolutely > straight line motion ALSO has angular momentum about an axis off that > path. How can I guarantee that it will be straight? Constrain the > object to glide on a rail -- I don't care. But in this case, the path > is *straight* and there is no center to find. Where would this *straight* rail be located? Planets are spherical, so it could not be located on a planet. Our current understanding of space, is that it is curved, so it could not be located in space either. So what does this tell you? > > > > Could you tell me what you think is the difference between angular > > > > momentum and spin? [quoted text clipped - 14 lines] > > PD > > [snip] > > > > [quoted text clipped - 51 lines] > point, unless it is affected by the gravity of some other object, which > will then alter that point which the comet is rotating around. Not necessarily! Ever hear of parabolic or hyperbolic orbit? http://en.wikipedia.org/wiki/Hyperbolic_trajectory http://www.go.ednet.ns.ca/~larry/orbits/ellipse.html Look at those links and then read cases (a)-(e) again. > >From wiki, > "Deflection is the physical event where an object collides and bounces [quoted text clipped - 4 lines] > > For me, the whole universe is rotating around its center point What center point? Where is it? How do we find it? > , so > everything in it is also rotating around a point. [quoted text clipped - 15 lines] > not be located in space either. > So what does this tell you? This points to a tangent (no pun intended), so I'll set this aside for now. > > > > > Could you tell me what you think is the difference between angular > > > > > momentum and spin? [quoted text clipped - 14 lines] > > > > PD > > > [snip] > > > > > [quoted text clipped - 57 lines] > > Look at those links and then read cases (a)-(e) again. Do you think the point the comets are rotating around must be in our solar system? Start thinking on larger scales and you will find that point of rotation. For cases d and e, they may leave our solar sytem, but do they leave our galaxy? Like I said, they are affected by the gravity of other objects. For us, here on little planet Earth, the trajectory might look like a straight line, but from the Milky Way's point of view it could be just a little tiny circular/elliptical orbit. The thing is, we both agree that they have angular momentum. > > >From wiki, > > "Deflection is the physical event where an object collides and bounces [quoted text clipped - 6 lines] > > What center point? Where is it? How do we find it? Haha. Why, the point at the center of the universe of course, which is at the center of the universe, which we find by travelling to the centre of the universe. And how do we do that you may ask. Oh, that's simple, we just put time in reverse for a few billion years. > > , so > > everything in it is also rotating around a point. [quoted text clipped - 37 lines] > > > > > > PD > Do you think the point the comets are rotating around must be in our > solar system? [quoted text clipped - 10 lines] > > The thing is, we both agree that they have angular momentum. I was going to explain that one does not need to find THE point that a body is rotating around for it to have angular momentum, but I don't think you'd believe me, so skip it. > > > >From wiki, > > > "Deflection is the physical event where an object collides and bounces [quoted text clipped - 12 lines] > And how do we do that you may ask. Oh, that's simple, we just put time > in reverse for a few billion years. I have news for you. There's no center. If there were, there'd be an edge to the universe. There's no edge. PD [snip] If there is no center, and no edge, wouldn't that imply there was no big bang? > If there is no center, and no edge, wouldn't that imply there was no > big bang? Not at all. The fact is, everyplace in the universe is like every other place in the universe in the sense that no matter where you go, everything is receding from that spot. Here's a little picture to help you get used to the idea. Imagine a ruler with inch marks on it, and it's expanding. Clamp one end down for a minute (we'll let go later) and let it expand. Say at 1", the 1" mark is creeping to the right because of the expansion. Let's suppose the creep rate that ends up being 0.1"/min at that mark. This means the 2" mark must be creeping to the right twice as much, since the space between 1" mark and 2" mark has to be growing too, and the 1" mark is sliding to the right already. OK, so the 2" mark is creeping to the right at 0.2"/min. Naturally, the 3" mark is creeping to the right at 0.3"/min and the 8" mark is creeping to the right at 0.8"/min for the same reason. OK so far? Next step: Notice that if you were to sit on the 3" mark as it slid to the right, the 2" mark and the 4" mark would both be receding from you -- the 2" mark is receding because it isn't moving to the right as fast as you are, and the 4" mark is receding because it's moving to the right faster than you are. In fact, if you think about it a second, if you were sitting on the 3" mark, both the 2" mark and the 4" mark would appear to be receding from you at a rate of 0.1"/min, right? In fact, if you were sitting on the 8" mark, the 7" mark and the 9" mark would both appear to be receding from you at 0.1"/min. So, except for marks near the end of the rulers, every position on the ruler appears identical, because your nearest neighbors are receding from you at 0.1"/min. Still with me? Next step: Make sure no marks are near the ends of the ruler, by making the ruler very, very long. (Infinitely long, if you want, in *both* directions. It's ok to let go of the ruler now, unclamp it.) It is *still* true that at every point on the ruler, the nearest neighbors appear to be receding at 0.1"/min. Now every mark on the ruler appears the same as any other mark, because they all see this very same neighbor behavior. Next step: If we can imagine an expanding ruler, surely we can start over and imagine a shrinking ruler. Just run it backwards in time, if you want. So suppose the 1" mark is moving to the *left* (toward 0) at 0.1"/min now. It will obviously take 10 min for that 1" mark to collapse to the 0" mark next to it. Now recall, the 2" mark was traveling at 0.2"/min, so now it's moving toward 0" at 0.2"/min, and it will take 10 min for that 2" mark to move 2" to sit on top of the 0" mark. This means that the 1" mark and the 2" mark will both collapse onto the 0" mark after 10 minutes; they will arrive at the same time. And likewise, by running this backwards, we find that *every* point on the ruler will collapse to 0 at the same time (after 10 minutes). Or rather, every point will collapse onto its neighbor after 10 minutes -- all the points will be on top of each other. Now think about this a second. This means that the whole *infinite* line will collapse so that all the points are on top of each other after 10 minutes of clock-back. That's the Big Bang. Final step: Repeat the same clock-back, but erase all the numbers from the inch marks -- they're not necessary. Since all marks collapse onto their neighbor marks after 10 minutes, we don't know if we've collapsed onto the zero mark or some other mark -- and it doesn't matter. All we know when we do the clock back is: - all marks on the ruler look identical to each other - every mark will collapse to its nearest neighboring marks after 10 minuts, on both sides, which of course means that they'll all collapse on top of each other at the same time (10 minutes). - there is no end to this infinite ruler, no 1-dimensional "edge" if you will. - there is no recognizable *center* anymore. All the marks are moving *relative* to each other, there's no external reference to say one of them is not *absolutely* moving. *No matter where* I sit on the ruler, the whole infinite ruler is going to collapse onto where I am in 10 minutes. Now, instead of a one-dimensional ruler, do the same in three space dimensions (and one time dimension), and you have the Big Bang. - Infinite universe, no edge (no ends to the ruler in any direction) - All points in the universe are identical - Universe collapses in a finite time to an infinitely dense singularity There is no need for an edge to have a big bang, and thus there is no need for a center. Sketches as a function of time (1 min, 2 min, etc.) will help. PD Do you mind if I point out a logical fallacy in your description? > > If there is no center, and no edge, wouldn't that imply there was no > > big bang? <snipped> > Final step: > Repeat the same clock-back, but erase all the numbers from the inch [quoted text clipped - 27 lines] > > PD While this does an excellent job of explaining how and why scientists claim there's no center to the universe, it also reveals a flaw in either their logic or the description of the Big Bang event. At T=0 seconds, all points on the ruler are at the same place and are essentially the same point. At T=1 qt into the expansion (assuming time is quantized, 1 qt is the quantum time unit), all of the ruler points are now separated and no longer the same point if you follow the methodology mentioned in your post. Now here's where the logic gets strange. If we are capable of noticing from inside this expanding universe that it is in fact expanding, then the size of the universe is not infinite. Why? Well, put mathematically, for all Inf that is an infinity of the same order, Inf + 1 = Inf - 1 This means that the size of the universe at T=1qt is the same as it would be at T=2qt or T=100qt or T=15 billion years, regardless of the rate of expansion. As such, if the universe were expanding and it is infinite in size, then our units of measure would be expanding along with the universe and we would not notice any change. The fact that we can measure the expansion rate means that the universe is in fact finite in size. A more likely truth is that the center of the universe is not within the universe itself. Consider the 3D+1 universe like a balloon. If you consider the surface of a balloon as a 2D surface and we posit that the 3rd dimension is time, then the surface of the balloon represents the observable universe. At T=0 seconds, the entire surface of the universe is a mere point. At T=1qt, the universe now has a surface area based on the rate of radial expansion (which can be calculated from the linear expansion we observe). The point at the center of the universe now only exists in the past. The long and short of this idea is that the universe as we understand it exists on the surface of an expanding 4D hypersphere. This would explain the curvature of space, and why we see quasars at the edges of the visible universe. Since gravity would represent a dent in the surface, this would explain gravity's effect on time... but I'm getting away from the point. The universe does not have to be infinitely large for us to not be capable of locating it's center. But if the universe is indeed expanding, it MUST be expanding away from somewhere. That does not imply that this somewhere can be found within the limits of our observable universe. R. > If there is no center, and no edge, wouldn't that imply there was no > big bang? No. Why do you think so? The Big Bang was *not* an explosion happening at one point, with matter and energy spreading away from that point spherically in all directions. Bye, Bjoern > > If there is no center, and no edge, wouldn't that imply there was no > > big bang? > > No. Why do you think so? The the universe started from a point and then expanded, then logically, the point it expanded from is the center. > The Big Bang was *not* an explosion happening at one point, with > matter and energy spreading away from that point spherically in all > directions. Then what was it? [snip] >>What center point? Where is it? How do we find it? > [quoted text clipped - 3 lines] > And how do we do that you may ask. Oh, that's simple, we just put time > in reverse for a few billion years. And why do you think this will lead us to the center of the universe? [snip] Bye, Bjoern > [snip] > [quoted text clipped - 7 lines] > > And why do you think this will lead us to the center of the universe? If the universe expanded from a point, then by reversing time, it would contract to a point, which would show us where the center of the universe is. [snip] > For me, the whole universe is rotating around its center point, > so everything in it is also rotating around a point. What makes you think that 1) the universe has a center and 2) everything in it rotates around it? > If this is not the case, then galaxies are rotating around a point, and > so, everything in a galaxy is rotating around a point. > > Everything has angular momentum! Probably, yes. [snip] > Our current understanding of space, is that it is curved, Wrong. Our current understanding of space is that it is (at least on large scales) essentially flat. [snip] Bye, Bjoern > [snip] > [quoted text clipped - 4 lines] > 1) the universe has a center and > 2) everything in it rotates around it? Solar systems have centers and objects in the solar systems rotate around those centers. Galaxies have centers and objects in those galaxies rotate around the center. So, why should the universe be any different? > > If this is not the case, then galaxies are rotating around a point, and > > so, everything in a galaxy is rotating around a point. [quoted text clipped - 9 lines] > Wrong. Our current understanding of space is that it is (at least > on large scales) essentially flat. According to Einstein, gravity is spacetime curvature, so therefore, masses will curve spacetime. Since gravity obeys the inverse square law, there will not be any place in spacetime that is not curved. Hence, space is curved. > [snip] > > Bye, > Bjoern [snip] >>Your subject line should have been more like: >>"New theory shuffles things around so that fermions have odd spin and >>bosons have even spin - no big deal, but it's prettier to me." > > That would be too long to fit on screen. But true. [snip] >>That's why experiment and the theoretical underpinnings are more >>important than numerology. > > Is there any experiment that can prove that fermions have half-intger > spin and bosons integer spin, as opposed to fermions having odd integer > spin and bosons even integer spin? If one defines spin like one usually defines angular momentum, then yes. > What are the theoretical underpinnngs you are talking about here? Quantization of angular momentum, following from the basic commutation relations. > Where does numerology come into it? In your attempt to rewrite the equations by merely using the identity hbar = 2 hdot. [snip] >>That of course is irrelevant. Two things do not have to be the very >>same object to be classified as the same kind of thing. Here it's >>important to understand what makes a class of things a species. > > They may both be a type of angular momentum, but they behave > differently. In what way do spin and orbital angular momentum behave differently? [snip] >>>Look, angular momentum is a vector, spin is a spinor. Why can you not >>>see that they are diferent? [quoted text clipped - 13 lines] > around a point, while spin is not associated with any rotation. Is this > correct? No, that's only right classically. Angular momentum in QM is essentially defined by its commutation relations and its behaviour under rotations of the coordinate system. > Could you tell me what you think is the difference between angular > momentum and spin? 1) Their g-factors are in general different. 2) The orbital angular momentum quantum number can only be an integer (look up the reason yourself), while the spin quantum number can also be an integer multiple of 1/2. Bye, Bjoern > [snip] > [quoted text clipped - 16 lines] > > If one defines spin like one usually defines angular momentum, then yes. What is it then? > > What are the theoretical underpinnngs you are talking about here? > [quoted text clipped - 16 lines] > > In what way do spin and orbital angular momentum behave differently? Why ask me a question I have only just answered, and which you have just answered yourself. Do you really want to know if I can cut & paste? > [snip] > [quoted text clipped - 28 lines] > (look up the reason yourself), while the spin quantum number can also > be an integer multiple of 1/2. Or in my model, integer multiples. > Bye, > Bjoern [snip] > And that's where you're wrong. Classical angular momentum and spin are > both pseudovectors (sometimes called spinors). Pseudovectors are sometimes called spinors? Would be news to me. [snip] Bye, Bjoern > [snip] > [quoted text clipped - 7 lines] > Bye, > Bjoern Would be to me, too. I don't know what I thought I was saying. PD >>Your calculations show nothing of the sort. What you said is: >>IF the expression for the expectation value for the total spin is [quoted text clipped - 4 lines] > > That is what I have been saying. No, that's a far weaker statement than your categorical statement "I have shown that the spin is 2". >>>>But it's also natural for something to have a spin s = 1/2. In this >>>>case, ms can have values -1/2 and 1/2, separated by a step of 1, you [quoted text clipped - 12 lines] > all integers, why should s be any different? My method fits in with the > existing data perfectly. As I already pointed out: your method would imply that l can only be multiples of 2. Or do you want to treat l and s differently? That also would be inconsistent. >>>My calculations show that if there are 2 states, it is spin 1, which is >>>natural. [quoted text clipped - 13 lines] > No, I i based this on the fact that spin can only be multiples of > hbar/2. Do you mean the magnitude or the z component here? > If a 50 year old person is asked his age, does he say that he is 100/2? > Of course not! No. > This is why I created a new constant and called it hdot. What on earth has the age of someome to do with spin? That's one of the strangest analogies I've ever seen. > hbar/2 is nonsense. Why? > I don't know what your'e getting at here. You get the same number of > spin projection states from either equation. But with your equation, we get a lot of inconsistencies elsewhere, for example in the treatment of orbital angular momentum. > I don't know much about spectrometry, so I can't comment on this at the > moment. Well, why don't you educate yourself first on a topic before attempting to change some of its fundamental equations? [snip] > Look, angular momentum is a vector, spin is a spinor. Err, no. Spin is also a vector. Why do you think otherwise? > Why can you not see that they are diferent? Because they aren't? Bye, Bjoern In sci.physics.particle Bjoern Feuerbacher <bjoern.feuerbacher@pci.uni-heidelberg.de> wrote: :> If a 50 year old person is asked his age, does he say that he is 100/2? :> Of course not! : : No. Perhaps you should phrase the argument in the same terms as GB: The question is not "does a 50 year old say that he is 100/2." Rather, the question that you are being asked is analogous to asking why a 50 year old person reports his age as 5 x 10 rather than as (6 x 8) + 2? ----- Richard Schultz schultr@mail.biu.ac.il Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel Opinions expressed are mine alone, and not those of Bar-Ilan University ----- "Gentlemen, Ciccolini here may look like an idiot, and talk like an idiot, but don't let that fool you -- he really is an idiot." > >>Your calculations show nothing of the sort. What you said is: > >>IF the expression for the expectation value for the total spin is [quoted text clipped - 7 lines] > No, that's a far weaker statement than your categorical statement "I > have shown that the spin is 2". But I did show that the spin is 2, and I did say bosons and fermions needed redefining. > >>>>But it's also natural for something to have a spin s = 1/2. In this > >>>>case, ms can have values -1/2 and 1/2, separated by a step of 1, you [quoted text clipped - 16 lines] > multiples of 2. Or do you want to treat l and s differently? That also > would be inconsistent. I just recently found out that mathmaticians who study lie groups tend to use a normalization with steps of 2, so maybe I will apply this to l as well. And, I want to get rid of n. > >>>My calculations show that if there are 2 states, it is spin 1, which is > >>>natural. [quoted text clipped - 15 lines] > > Do you mean the magnitude or the z component here? z component. > > If a 50 year old person is asked his age, does he say that he is 100/2? > > Of course not! [quoted text clipped - 5 lines] > What on earth has the age of someome to do with spin? That's one of > the strangest analogies I've ever seen. Nothing at all. > > hbar/2 is nonsense. > > Why? What would you rather say and write, hbar/2 or hdot. > > I don't know what your'e getting at here. You get the same number of > > spin projection states from either equation. [quoted text clipped - 7 lines] > Well, why don't you educate yourself first on a topic before attempting > to change some of its fundamental equations? Ha, that's a good one. Since I dont know about spectrometry, how could I poosibly know I was changing any of its fundamental equations? > [snip] > [quoted text clipped - 8 lines] > Bye, > Bjoern [cut] >"Whereas classical angular momentum arises from the rotation of an >extended object, spin is not associated with any rotating internal >masses, but is intrinsic to the particle itself." > >Therefore, 'intrinsic angular momentum' and 'angular momentum' are not >the same. Two questions. 1. How many spin states are possible according to current theory? IOW, what is the maximum number of spin states? By 'spin', I mean 'intrinsic angular momentum', a discrete intrinsic property of a fundamental particle. 2. Is there a theory or hypothesis out there that associates intrinsic spins with the dimensionality of the universe? Just curious. Louis Savain The Silver Bullet: Why Software Is Bad and What We Can Do to Fix it http://users.adelphia.net/~lilavois/Cosas/Reliability.htm > [cut] > >"Whereas classical angular momentum arises from the rotation of an [quoted text clipped - 10 lines] > 'intrinsic angular momentum', a discrete intrinsic property of a > fundamental particle. I don't think anyone knows. You would have to total the spins of all the particles in the universe. > 2. Is there a theory or hypothesis out there that associates intrinsic > spins with the dimensionality of the universe? [quoted text clipped - 5 lines] > The Silver Bullet: Why Software Is Bad and What We Can Do to Fix it > http://users.adelphia.net/~lilavois/Cosas/Reliability.htm [turgid gobble] Seriously there GB. You should do some reading and studying of these matters before you post any more. Your notions are grossly misinformed and make you look like a public school student mouthing cracked crap. You still don't have, just as an example, any clue about significant digits. So you mark yourself as a fool within the first couple lines of your post. If you would like some suggestions on introductory books to read I can suggest a few. Sadly, you are going to need to start at the grade 10 level or so. You have a great deal of remedial work to do before you can even understand the questions, never mind answering them. Socks >[turgid gobble] > [quoted text clipped - 13 lines] >answering them. >Socks I just love it when ignoramuses suggest that others do some reading. Let me ask you this, Sock-head. Do you know why things fall or why they move? Do you know why paricles have intrinsic spins? Do you know why an electron has an electric field around it? Answer: No. I would suggest that you do some reading but, since you won't be able to find any of this information anywhere (physicists are all a bunch of ignoramuses), all I can suggest is for you to go do some thinking of your own. You might discover something new, Louis Savain The Silver Bullet: Why Software Is Bad and What We Can Do to Fix it http://users.adelphia.net/~lilavois/Cosas/Reliability.htm I am reading and I am studying these matters. Instead just of saying my notions are grossly misinformed, point out to me where you think they are misinformed. Whats your big deal with significant digits, hdot = h / 4pi, pi has been calculated to over a biliion decimal places. So hdot could probably be calculated for 1 decimal place to about a billion. Everyone knows what significant digits are, well at least they should. Your assumptions that other people don't know what they are doing or talking about, just proves that you are an arrogant turd. Look, I'm not writing these equations down on paper, I cut the answer from my calculator and paste it into my document, so why would I be bothered with significant digits, the values are still the same basically. And you call me a fool! Erm, what's grade 10 level. See, your assuming that I'm from your country, which I guess I'm not since I dont know what grade 10 is, and I couldn't be bothered checking. Let me make an asumption here. Your'e a Yank! > I am reading and I am studying these matters. > [quoted text clipped - 3 lines] > Whats your big deal with significant digits, hdot = h / 4pi, pi has > been calculated to over a biliion decimal places. Err, but h is known only to very few digits precisely. > So hdot could > probably be calculated for 1 decimal place to about a billion. Why on earth do you think so? Don't you know that the result of a calculation has *at most* so many significant digits as the number with the *least* significant digits in the calculation? > Everyone > knows what significant digits are, well at least they should. Apparenlty you don't. > Your > assumptions that other people don't know what they are doing or talking > about, just proves that you are an arrogant turd. Well, you have just shown above that you indeed don't know how to work with significant digits... > Look, I'm not writing these equations down on paper, I cut the answer > from my calculator and paste it into my document, so why would I be > bothered with significant digits, the values are still the same > basically. > > And you call me a fool! Well, the paragraph you just wrote shows that you are indeed one. Your attitude is totally naive. [snip] Bye, Bjoern > > I am reading and I am studying these matters. > > [quoted text clipped - 5 lines] > > Err, but h is known only to very few digits precisely. So what? > > So hdot could > > probably be calculated for 1 decimal place to about a billion. > > Why on earth do you think so? Don't you know that the result of a > calculation has *at most* so many significant digits as the number > with the *least* significant digits in the calculation? Do you reckon? Bang the numbers into your calculator and presto, look at that, now there's well more digits. > > Everyone > > knows what significant digits are, well at least they should. > > Apparenlty you don't. Why? Because I just cut and paste from my windows calc? > > Your > > assumptions that other people don't know what they are doing or talking > > about, just proves that you are an arrogant turd. > > Well, you have just shown above that you indeed don't know how to work > with significant digits... No, what I have shown is that I could not be bothered. Have you any idea how many equations are in my document? No! Have you any idea how many values are in my document? No! > > Look, I'm not writing these equations down on paper, I cut the answer > > from my calculator and paste it into my document, so why would I be [quoted text clipped - 5 lines] > Well, the paragraph you just wrote shows that you are indeed one. > Your attitude is totally naive. It shows no such thing. But what you just said shows that you are a arrogant fool and you think you are superior to other people. > > > I am reading and I am studying these matters. > > > [quoted text clipped - 7 lines] > > So what? So, this is an important point. It is scientifically improper to declare knowledge of digits you don't have. If I measure a mass to be 15.3 grams and a volume to be 3.7 mL, my calculator will tell me a density of 4.1351351351351 g/mL, but it would be lying to me because I only know it to be 4.1 g/mL. I know nothing about any digits beyond that. This is one of the reasons why numerology is distasteful -- there's no numerical literacy in it. PD > > > So hdot could > > > probably be calculated for 1 decimal place to about a billion. [quoted text clipped - 36 lines] > It shows no such thing. But what you just said shows that you are a > arrogant fool and you think you are superior to other people. hdot = 5.2728584e-35 m^2.kg.s^-1 satisfied now. hdot = 5.2728584e-35 m^2.kg.s^-1 satisfied now. hdot = 5.2728584e-35 m^2.kg.s^-1 satisfied now. >>>Remember this, >>> [quoted text clipped - 3 lines] > > I have answered this before, but will do so again. Apparently I missed that, sorry. > A quantum is a discrete indivisible value, the smallest value something > can have. Well, since the smallest value which |S| or |L| can have is *not* hbar, so hbar is not the quantum of angular momentum, according to your own definition. > For example, if something has a quantum of 1, it will only be allowed > to have a value which is a multiple of 1. Well, but |S| and |L| are not integer multiples of hbar. > It can not have a value of 0.5 or 1.5. How about |L| = sqrt(2) hbar? > If you disagree with this, then tell me what you think it means. First tell me how you can call hbar the quantum of angular momentum in the light of the facts above. >>>spin is not angular momentum, >> [quoted text clipped - 4 lines] > extended object, spin is not associated with any rotating internal > masses, but is intrinsic to the particle itself." Indeed. > Therefore, 'intrinsic angular momentum' and 'angular momentum' are not > the same. Non sequitur. "is not associated with any rotating internal masses" is not equivalent to "is not intrinsic angular momentum". Bye, Bjoern > >>>Remember this, > >>> [quoted text clipped - 12 lines] > hbar, so hbar is not the quantum of angular momentum, according to > your own definition. I never even mentioned |S| or |L| and I certainly didn't define hbar as the quantum of angular momentum. Where are you getting this from? > > For example, if something has a quantum of 1, it will only be allowed > > to have a value which is a multiple of 1. > > Well, but |S| and |L| are not integer multiples of hbar. I never said they were. > > It can not have a value of 0.5 or 1.5. > > How about |L| = sqrt(2) hbar? What about it? > > If you disagree with this, then tell me what you think it means. > > First tell me how you can call hbar the quantum of angular momentum > in the light of the facts above. In light of the facts above, I have no idea what you are talking about, as you appear to be making it up as you go along. > >>>spin is not angular momentum, > >> [quoted text clipped - 14 lines] > "is not associated with any rotating internal masses" is not > equivalent to "is not intrinsic angular momentum". Are you trying to say that spin is not intrinsic angular momentum? > hbar is the quantum of angular momentum > spin is not angular momentum And "spin 1/2" refers to 1/2 h-bar. Calling it 1 (h-bar/2) is as irrelevant as calling a rose a thorn and proclaiming that all roses are thorns. > > hbar is the quantum of angular momentum > > spin is not angular momentum > > And "spin 1/2" refers to 1/2 h-bar. Calling it 1 (h-bar/2) is as > irrelevant as calling a rose a thorn and proclaiming that all roses are > thorns. I dont call it 1 (hbar/2), I call it 1 hdot. What looks better to you? > We have all noticed that spin is described as being a multiple of > hbar/2. I thought that it would be better to set this value to a > constant giving, Why? hbar occurs also in a lot of other formulas and quantities. It is not only connected with spin. Do you want to replace hbar with 2 hdot in all these equations, or what? > hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s > [quoted text clipped - 34 lines] > > Would these redefinitions have any other effects on the Standard Model? No, since what you did is mathematically completely equivalent to the standard rules. It amounts to something like a change of the unit system merely. > Can these redefinitions explain any currently unexplained phenomena? No, since all you did was to change the units. You didn't introduce any new physics. > Are there any experiments that could confirm or refute these claims? No, since your formulation is completely equivalent to the standard one. [snip] Bye, Bjoern > > We have all noticed that spin is described as being a multiple of > > hbar/2. I thought that it would be better to set this value to a [quoted text clipped - 3 lines] > It is not only connected with spin. Do you want to replace hbar > with 2 hdot in all these equations, or what? hbar is the quantum of angular momentum. hdot is the quantum of spin. Spin and angular momentum are not the same thing. For equations relating to angular momentum, hbar should be used. For equations relating to spin, hdot should be used. > > hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s > > [quoted text clipped - 38 lines] > the standard rules. It amounts to something like a change of the unit > system merely. That must of been an exhausting investigation, but I also *think* that this will be the case. > > Can these redefinitions explain any currently unexplained phenomena? > > No, since all you did was to change the units. You didn't introduce > any new physics. Once again, that must of been an exhausting investigation. You dont neccessarily need to introduce new physics to explain something, sometimes you just need to look at it from a different perspective. In this case, odd and even integer spins as opposed to half integer and integer spins. > > Are there any experiments that could confirm or refute these claims? > > No, since your formulation is completely equivalent to the standard one. And for the last time, that must of been an exhausting investigation. It took you what, 2 hours? I agree that it shouldn't, but you never know. > We have all noticed that spin is described as being a multiple of > hbar/2. I thought that it would be better to set this value to a [quoted text clipped - 49 lines] > It may also be of interest to note that the only even prime number is > 2. There are so many errors in this post that I hardly know where to begin. PD Anywhere you like > Anywhere you like My mistake. I thought you were claiming something you apparently are not. My apologies. I will, however, note that you've traded one "unnaturalness" for another. Now you s values are integers, but your spin projection quantum numbers have to increment in steps of two rather than one. PD > > Anywhere you like > [quoted text clipped - 6 lines] > > PD I know, it has been bugging me that, so now I have to ask, where does the 1 come from in the equation: |sv| = sqrt(s(s + 1)) * hbar > > > Anywhere you like > > [quoted text clipped - 11 lines] > > |sv| = sqrt(s(s + 1)) * hbar Ah, for that, it's time to pick up that copy of Eisberg and Resnick I was telling you about. You won't find it there, but you'll need to read that before you read Griffiths. PD Mathematicians who study lie groups often use a normalization with steps of two. > We have all noticed that spin is described as being a multiple of > hbar/2. I thought that it would be better to set this value to a > constant giving, [snip crap] Idiot.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > > We have all noticed that spin is described as being a multiple of > > hbar/2. I thought that it would be better to set this value to a [quoted text clipped - 8 lines] > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf This guy is a pure genius to all us mere idiots. Uncle Al, do yourself a favour, get shut of that wbesite or at least try and learn how to build a decent one. It looks like it was made by a 10 year old on drugs. I just went through the equations for the fundamental constants listed on NIST, and found that it is beneficial to replace hbar with hdot in the following equations. Bohr magneton, muB = e.hbar / 2.me = e.hdot / me Nuclear magneton, muN = e.hbar / 2.mp = e.hdot / mp Gyromagnetic ratio, Yx = 2.|mux| / hbar = |mux| / hdot, where x is a specific particle. This suggests that hdot is related to magnetism. Will anyone tell me some equations that have h or hbar in them? I forgot to add this one, pP = hdot / lP = where hdot = 5.2728584e-35 m^2.kg.s^-1 lP = Planck length = 1.61624e-35 m pP = 3.2624229e m.kg.s^-1 Does this number look familiar? It should do! >From wikipedia, "By the Heisenberg uncertainty principle of standard quantum mechanics, an object whose position was accurate to the Planck length would have an uncertainty in momentum approximately 3.2629 kg m / s." I posted my findings to wiki Talk:Fermion, and got the following response. "The change you're proposing is a trivial renormalization; physicists are perfectly happy with the convention as it stands, tho mathematicians studying lie groups tend to use a normalization with steps of two." Can anyone explain in simple terms whre this "steps of 2" comes from. I just used steps of 2 to get the correct results, without knowing any of this. > I posted my findings to wiki Talk:Fermion, and got the following > response. [quoted text clipped - 8 lines] > I just used steps of 2 to get the correct results, without knowing any > of this. You've just stumbled on the different choices of rescaling the parameters. Physicists choose the one they do because the physics laws that they use to derive the spin quantum numbers end up with the step size of one and the possibility of half-integer spin. Mathematicians come at it from a different direction and they make a different choice. The key word in the response above is "trivial". PD I have just noticed a mistake with my oroginal post, where I said: For spin 2 particles this give: |sv| = sqrt(8).hdot and Sz = -hdot, 0, +hdot This should be: For spin 2 particles this gives: |sv| = sqrt(8).hdot and Sz = -2.hdot, 0, +2.hdot |
Reply to this Message |
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2009 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.