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Natural Science Forum / Physics / Particle Physics / October 2005Posted another Preview Paper
Hello again: I just today posted a second preview paper at http://home.nycap.rr.com/jry/FermionMass.htm. This paper actually previews perhaps four or five distinct papers that I am considering for future publication. These include the relationship between duality symmetry and the geometrodynamic vacuum, how to deal with large running couplings outside the range of perturbation theory, electroweak / strong and leptoquark unification, and superconductivity. They are all related to my two papers at http://arxiv.org/abs/hep-ph/0509223 and http://arxiv.org/abs/hep-ph/0508257 dealing with duality symmetry and electric and magnetic charges. The website contains an introduction which can help you navigate through. Comments and suggestions are appreciated. Jay. _____________________________ Jay R. Yablon Email: jyablon@nycap.rr.com > Hello again: > [quoted text clipped - 18 lines] > Jay R. Yablon > Email: jyablon@nycap.rr.com Hi Jay, studied your site. 3rd rank asymmetrical tensors are fun... Probably the most famous 3rd rank asymmetric tensor is Maxwell's 0 = F_ab,c + F_bc,a + F_ca,b and describes an EM-wave, which is a typical boson. ("," is a partial and ";" is a covariant later). 1) We can define a 2nd rank tensor, A_a B_b = - A_b B_a and specify (A_a B_b);c =0 , (a symmetry and conservation) and then find, A_a;c B_b = - B_b;c A_a is that asymmetric in indices "a,b,c" ? and symmetric if A and B are exchanged? 2) Set (A_a A_b);c =0 and find A_a;c A_b = - A_b;c A_a and see an asymmetry in indices "a" and "b", but does it follow "c" and "a" or "b" are asymmetric? In problem (1) the vectors A and B are distinct, so one is applying "relational" GR. In problem (2) vector "A" can exist on a continuum, as it is not relative to another vector so it can apply to a point. I wonder, do the mathematicians have all this stuff figured out in catagories or something. I'm thinkin' if your moving tensors into particle physics you may as well have lots of ammo, IOW's know your descriptive options. The problems above are parallel to my interest, so perhaps we could work out a scheme if one is not available for applications to GR and particle physics. Best Regards Ken S. Tucker > Hi Jay, studied your site. > [quoted text clipped - 4 lines] > > 0 = F_ab,c + F_bc,a + F_ca,b Ken that is THE tensor on which this is ALL based. Not just an example; at the heart of what I am doing. A major thrust of http://arxiv.org/abs/hep-ph/0508257, which I know you are studying, is to show that this tensor is NOT equal to zero when one imposes a LOCAL duality symmetry (broken or unbroken). And the large paper I just posted today shows also how this tensor becomes NON-ZERO for non-Abelian interaction, such as weak and strong. So, the question becomes, if this "most famous 3rd rank asymmetric tensor" should ever become non-zero (and I show two ways this can happen, separately and in combination) and represent a real, observable, physical object, what would that object be seen as? My answer: a baryon! When Maxwell wrote 0 = F_ab,c + F_bc,a + F_ca,b (of course he didn't use tensors at the time and didn't know about spacetime and so had two equations for this), Maxwell had discovered a baryon, but his baryon was equal to zero. It has taken 130 years for us to recognize that Maxwell is the one who found the first, very tenuous and hidden clues, to the existence of three-quark baryons. That is the point I wish to make here. I do not have a full proof of this, but I will just about bet the farm that people will eventually come to understand that the covariant object for a baryon is a third rank antisymmetric tensor T_abc = F_ab,c + F_bc,a + F_ca,b when this tensor is NOT equal to zero. And, that a deuteron, which is one example of two baryons "glued" together via strong interactions, involves two of these third rank antisymmetric tensors contracted together along at least one index, maybe more, possibly all three. That is, deuteron = T^abc T_abc if the contraction is all three indexes. If some of the indexes remain free, then other contractions are possible. Understanding this, will also give us a much cleaner understanding of inter-nucleon strong interactions, which is where nuclear physics was at about 40 years ago before Gell Mann and others came up with quarks. But, of course, this will be on a foundation in which each nucleon is a three quark object T_abc, per the Feynman diagram I just posted. I am curious how to construct, say, an entire hydrogen atom, deuteron plus electron electric current J^u (and magnetic current P^u following http://arxiv.org/abs/hep-ph/0509223) out of tensor objects, following this approach. And even larger, more complex atoms, all out of tensors. Jay. Hi Jay, scrape me off the ceiling! I'll do a bit a bit of inline... > > Hi Jay, studied your site. > > > > 3rd rank asymmetrical tensors are fun... > > > > Probably the most famous 3rd rank asymmetric tensor is > > Maxwell's 0 = F_ab,c + F_bc,a + F_ca,b Eq(0) > Ken that is THE tensor on which this is ALL based. Not just an example; at > the heart of what I am doing. [quoted text clipped - 9 lines] > and in combination) and represent a real, observable, physical object, what > would that object be seen as? My answer: a baryon! Ok, for brevity let me redo Eq.(0) as ExB => providing direction C. In a vacuum (no fields) C=c, however near matter (aka baryons, or any mass) C =/= c, because light deflects to indicate the presence of matter, equivalent in that case to, T_abc =/= 0 (T_abc is introduced below), and ExB = C =/= c because of deflection. (Jay I'm working to keep GR and EM connected here). > When Maxwell wrote 0 = F_ab,c + F_bc,a + F_ca,b (of course he didn't use > tensors at the time and didn't know about spacetime and so had two equations > for this), Maxwell had discovered a baryon, but his baryon was equal to > zero. It has taken 130 years for us to recognize that Maxwell is the one > who found the first, very tenuous and hidden clues, to the existence of > three-quark baryons. That is the point I wish to make here. Yup so far. > I do not have a full proof of this, but I will just about bet the farm that > people will eventually come to understand that the covariant object for a [quoted text clipped - 10 lines] > be on a foundation in which each nucleon is a three quark object T_abc, per > the Feynman diagram I just posted. Tough problem. Suppose the Baryon is a rank 3 tensor, and that looks reasonable, now you next moved to Baryon inter- action, of which two are available, 1) Tensor summation, corresponding to the boson bonding nature. 2) Tensor multiplication, that includes Pauli's exclusion principle. In my understanding you are clear to use summation with bosons all over the field because boson-boson reactions are nil. OTOH, the fermion nature of the Baryons (or Leptons) T_abc would need a product for Pauli's exclusion, because it's a relation, (musical chairs). > I am curious how to construct, say, an entire hydrogen atom, deuteron plus > electron electric current J^u (and magnetic current P^u following > http://arxiv.org/abs/hep-ph/0509223) out of tensor objects, following this > approach. And even larger, more complex atoms, all out of tensors. Me too Jay, Fred was kind enough to reformat an article we prepared you may want to glance at, http://www.vacuum-physics.com/KST/GR_Charge_Couple3.pdf as a foundation. We're quite parallel in thinking. Ken S. Tucker Hi Jay, studied your site. 3rd rank asymmetrical tensors are fun... Probably the most famous 3rd rank asymmetric tensor is Maxwell's 0 = F_ab,c + F_bc,a + F_ca,b and describes an EM-wave, which is a typical boson. ("," is a partial and ";" is a covariant later). 1) We can define a 2nd rank tensor, A_a B_b = - A_b B_a and specify (A_a B_b);c =0 , (a symmetry and conservation) and then find, A_a;c B_b = - B_b;c A_a is that asymmetric in indices "a,b,c" ? and symmetric if A and B are exchanged? 2) Set (A_a A_b);c =0 and find A_a;c A_b = - A_b;c A_a and see an asymmetry in indices "a" and "b", but does it follow "c" and "a" or "b" are asymmetric? In problem (1) the vectors A and B are distinct, so one is applying "relational" GR. In problem (2) vector "A" can exist on a continuum, as it is not relative to another vector so it can apply to a point. I wonder, do the mathematicians have all this stuff figured out in catagories or something. I'm thinkin' if your moving tensors into particle physics you may as well have lots of ammo, IOW's know your descriptive options. The problems above are parallel to my interest, so perhaps we could work out a scheme if one is not available for applications to GR and particle physics. Best Regards Ken S. Tucker > Hello again: > [quoted text clipped - 18 lines] > Jay R. Yablon > Email: jyablon@nycap.rr.com > 1) We can define a 2nd rank tensor, > A_a B_b = - A_b B_a Before you get too carried away with what you can do with this idea, consider the following. Choose some basis, and consider the components A_a, B_b, a,b=1..n (presumably n=4, but it doesn't make any difference). A_0 B_0 = -A_0 B_0, so at least one of A_0 and B_0 must be 0. Suppose that A_0 is not 0. Then B_0 must be 0. Now, A_0 B_1 = -A_1 B_0 = 0, so B_1 must also be 0. Next, A_0 B_2 = -A_2 B_0 = 0, so B_2 must also be 0. And so on, so B_b must be the 0 vector, and so the tensor A_a B_b is in fact the 0 tensor. (Of course, if we suppose that B_0 is not 0, we deduce that A_a must be the 0 vector, which leads to the same conclusion.) Note: there is no tensor analysis involved in this argument whatever, so fiddling about with the metric or the definition of covariant derivative won't help. It's just algebra. You can't make a non-trivial skew tensor this way. As a general suggestion, whenever you're tempted to consider a particular construction, try to do an example, just to see if it works. It can save a lot of wasted effort. Thank you for your kind reply Mr. Low. > > 1) We can define a 2nd rank tensor, A_a B_b = - A_b B_a (0) > Before you get too carried away with what you > can do with this idea, consider the following. I studied your post carefully, let me rebut this way... A_a B_b = 1/2 (A_a B_b - A_b B_a) (1) + 1/2 (A_a B_b + A_b B_a) (2) Obviously I conditioned (2) to be zero, meaning my original tensor (0) is equivalent to (1), but abbreviated, but (1) is clearly asymmetrical. Nonsymmetrical tensors are tricky. Mr. Low attempted a solution using absolute values, which is not a good general prodecure. Consider the following physics application, Let F^ab be the EM field tensor and permit the following time and space relations, A_0 = B_0 and A_1 = -B_1 then, F^ab A_a B_b =/= 0 , = F^01 A_0 B_1 + F^10 A_1 B_0 = F^10 A_0 A_1 + F^10 A_1 A_0 = 2*F^10 A_1 A_0 , set F^10 = E = q/r^2, r = A_1, ct = A_0 , r=ct then = 2q , which is a nonzero invariant, the fundamental charge. Regards Ken S. Tucker > Choose some basis, and consider the components > A_a, B_b, a,b=1..n (presumably n=4, but it [quoted text clipped - 28 lines] > to do an example, just to see if it works. > It can save a lot of wasted effort. >>>1) We can define a 2nd rank tensor, > A_a B_b = - A_b B_a (0) >>Before you get too carried away with what you >>can do with this idea, consider the following. > I studied your post carefully, Study it again. It is no more than high school algebra. > let me rebut this way... > A_a B_b = 1/2 (A_a B_b - A_b B_a) (1) > + 1/2 (A_a B_b + A_b B_a) (2) Yes, this is true. But it doesn't help. > Obviously I conditioned (2) to be zero, meaning > my original tensor (0) is equivalent to (1), but > abbreviated, but (1) is clearly asymmetrical. Fine. So write out explicitly a pair of vectors A_a and B_b that has this property. (The elements of the empty set *all* have lots of interesting properties, but it doesn't do to spend too much of one's life studying them.) > Nonsymmetrical tensors are tricky. Not particularly. > Mr. Low > attempted a solution using absolute values, > which is not a good general prodecure. Mr Low gave you a simple algebraic argument that there are no two vectors A_a and B_b such that A_a B_b is skew and non-zero. But I do make mistakes, and it's conceivable that I made a mistake in this case. When you provide two non-zero vectors A_a and B_b such that A_a B_b = - A_b B_a, I will immediately concede that I made a mistake. Hi Robert, to less formal... > >>>1) We can define a 2nd rank tensor, > > A_a B_b = - A_b B_a (0) [quoted text clipped - 4 lines] > Study it again. It is no more than high school > algebra. Rob, please be careful, above you said, "Then B_0 must be 0." Now you CANNOT equate a component (tensor B_0) to an invariant "0" in mid-calculation. What you did is specialized the CS then proceeded to general conclusions. OTOH I maintained General Covariance to arrive, for example, at "2q". > > let me rebut this way... > > A_a B_b = 1/2 (A_a B_b - A_b B_a) (1) [quoted text clipped - 29 lines] > that A_a B_b = - A_b B_a, I will immediately > concede that I made a mistake. Yes, you maid a mistake, you snipped my physics example, I didn't snip you...anyway Mr. Low is right, I'd prefer to be Mr. Wrong. LOL Ken > Hi Robert, to less formal... >>>Ken S. Tucker wrote: [quoted text clipped - 13 lines] > > "Then B_0 must be 0." Yes, I said: pick a basis, and consider the components of A_a and B_b in that basis. Since A_0 B_0 = -A_0 B_0, the producet of the two components must be zero. Assuming that A_0 is not 0, B_0 must be 0. If you assume B_0 is not 0, then A_0 must be. Either way, A_a B_b ends up being identically 0. If a tensor is identically 0 in any basis, it is 0 in every basis. Therefore, if A_a B_b = -A_b B_a, it must be the 0 tensor. Unless, of course, you can exhibit a pair of vectors with the property that A_a B_b = - A_b B_a, without it being 0. > Now you CANNOT equate a component (tensor B_0) > to an invariant "0" in mid-calculation. I dind't do that. I argued that the component must be 0 *in whatever basis you picked*. The argument is that it then follows that *every* component of A_a B_b must be 0, and that therefore the tensor itself is 0. > What > you did is specialized the CS then proceeded > to general conclusions. Because a tensor that vanishes in any basis must vanish in all bases. > OTOH I maintained > General Covariance to arrive, for example, at > "2q". You might have arrived there, but you didn't start with a pair of vectors A_a and B_b such that A_a B_b is non-zero and skew. > Yes, you maid a mistake, you snipped my physics > example, It was irrelevant to the claim. > I didn't snip you... No, but you haven't responded to the question: can you find a pair of vectors with the property you claim? > Mr. Low is right, I'd prefer to be Mr. Wrong. Well, there's something I suspect we can agree on. Mr. Low, previously I proved from Eq.(0) we are working with the following tensor, 2*A_a B_b = A_a B_b - A_b B_a (1) is that clear? If so then we move to argue on the RHS of (1). Ref, A_a B_b = - A_b B_a (0) more below, > >>>>Before you get too carried away with what you > >>>>can do with this idea, consider the following. [quoted text clipped - 14 lines] > zero. Assuming that A_0 is not 0, B_0 > must be 0. Ok, let's go to dyads again, using two VECTORS "A" and "B" with my definition from Eq.(0) in dyadic form, AB = A.B + AxB = -BA with the dot "." being the scalar product and the "x" being the vector product, and see Spiegel's book, bottom of pg 73, where "dyads are generalized in tensor analysis". What Rob is doing is concluding from A.B=0 that A or B must be zero and that's wrong, he's making a mistake there, but to his credit a very good one! In tensor notation I show the AxB part of the dyadic appears as my Eq.(1) above. What Mr. Low has taught me is that we should take great care in going from Vector analysis into Tensor analysis, and we shouldn't skip over dyadics. Regards Ken S. Tucker > If you assume B_0 is not 0, then A_0 > must be. [quoted text clipped - 52 lines] > Well, there's something I suspect we can agree > on. > Mr. Low, previously I proved from Eq.(0) we are > working with the following tensor, > 2*A_a B_b = A_a B_b - A_b B_a (1) > is that clear? It's clear that you want it. It's also clear that it only happens when at least on of A and B is the 0 vector. > If so then we move to argue > on the RHS of (1). > > Ref, > A_a B_b = - A_b B_a (0) Look, just exhibit a pair of vectors that have that property. I repeat: if you think you can find two vectors which have that property, show me them. And now, on to the pointless bit. If arguing based on the components in a basis is too complicated for you, I strongly suspect that what I'm about to write will make as much sense to you as it would to a daffodil, but what the heck. Suppose that A_a B_b = - A_b B_a and that neither of A_a nor B_b is the zero vector. Then pick X^a such that neither A_a X^a nor B_a X^a is zero. It follows that B_b (A_aX^a) = - A_b (A_a X^a), so that A_a and B_b must be proportional, i.e. A_a = k B_a for some non-zero k. But now A_a B_b = k A_a A_b which is clearly symmetric. But we also require it to be skew. This can only happen if it is 0, and hence A_a is zero, which is a contradiction. > What Rob is doing is concluding from A.B=0 > that A or B must be zero and that's wrong, Except of course that it isn't. I'm concluding from the fact that the product of two real numbers (particular components of A and B in some basis) is 0 that one of the numbers must be 0, and that's right. Your inability to understand the relevance does rather baffle me, though. > he's making a mistake there, but to his > credit a very good one! Ooh, that's not the mistake I'm making. The mistake I'm making is that you're capable of following a simple mathematical argument. > What Mr. Low has taught me is that we should Alas, I fear I have taught you nothing. But I can always hope that at least the occasional lurker has been saved some effort. > > Mr. Low, previously I proved from Eq.(0) we are > > working with the following tensor, > > 2*A_a B_b = A_a B_b - A_b B_a (1) > > is that clear? > > It's clear that you want it. Good! Now what we want from you is to post the vector product AxB in tensor notation. Can you do that? Ken ... >>>Mr. Low, previously I proved from Eq.(0) we are >>>working with the following tensor, [quoted text clipped - 5 lines] > the vector product AxB in tensor notation. > Can you do that? No, no, after you. Give two vectors A and B such that A_a B_b = - A_b B_a (without it being the 0 tensor). I've given you two independent proofs that it can't be done. Each time you simply ignore the argument. You insist that it can be done, by 'conditioning' A and B such that A_a B_b - A_b B_a be 0. Of course, this is about as possible as arranging a square to have three sides by 'conditioning' one of the sides to be of length 0 without changing the others. But I'm always open to new ideas, so just show me an explicit example. > >>>Mr. Low, previously I proved from Eq.(0) we are > >>>working with the following tensor, > >>>2*A_a B_b = A_a B_b - A_b B_a (1) > >>>is that clear? > >> > >>It's clear that you want it. > > Good! Now what we want from you is to post > > the vector product AxB in tensor notation. > > Can you do that? > > No Ok then, we know what your problem is, how do you want to learn tensorized AxB? I read your posts Rob, but unless we can agree on AxB then we cannot proceed together. Post the question to sci.math, there's lots of smart guys in that group who might help you. Ken > Give two vectors A and B such that > A_a B_b = - A_b B_a (without it > being the 0 tensor). > I've given you two independent > proofs that it can't be [quoted text clipped - 11 lines] > But I'm always open to new ideas, > so just show me an explicit example. >>>>>Mr. Low, previously I proved from Eq.(0) we are >>>>>working with the following tensor, [quoted text clipped - 6 lines] >> >>No Oh, nice bit of dishonest quotation. And I notice you have *never* even responded to the point that you cannot find vectors A and B such that A_a B_b = -A_b B_a. Time to update that killfile. > I read your posts Rob, Looking at the words in them isn't the same as reading them. But I've invested all the effort I care to in trying to help you. You just carry on posting your tripe. > >>>>>Mr. Low, previously I proved from Eq.(0) we are > >>>>>working with the following tensor, [quoted text clipped - 13 lines] > find vectors A and B such that > A_a B_b = -A_b B_a. A0 = B0 , A1 = - B1 as explained! BTW, Low's solution of setting A0*B0 =0 means, -1 = 0/0 which is certainly not a proof. >>And I notice you have *never* even >>responded to the point that you cannot >>find vectors A and B such that >>A_a B_b = -A_b B_a. > A0 = B0 , A1 = - B1 Oh, I see. You're so clueless that you think that if A=[1,1] and B=[1,-1] then A^T B is skew-symmetric. Well, it was worth not kill-filing you until now just for that little gem. But don't worry. I won't annoy you any more by trying to get you to make contact with any form of mathematical coherence. Any further replies to your posts will just be in case any new lurkers might be led astray. > >>And I notice you have *never* even > >>responded to the point that you cannot > >>find vectors A and B such that A_a B_b = -A_b B_a. A0 = B0 , A1 = - B1 For brevity set C_ab = A_a B_b, and C_ab = -C_bc. Get the "norm" by C = AB = g^ab C_ab . IFF g^ab = s^ab (symmetrical) then C=0. However, IF g^ab = s^ab + a^ab (nonsymmetrical) then C=AB= a^ab C_ab =/=0. Indeed now g^ab C_ab = invariant =/=0, proving C_ab is not always zero. Mr. Low's failures were numerous, most importantly, subbing invariants into components. Regards Ken S. Tucker > Oh, I see. You're so clueless that > you think that if A=[1,1] and B=[1,-1] [quoted text clipped - 8 lines] > posts will just be in case any new lurkers > might be led astray. > A_a B_b = -A_b B_a. > A0 = B0 , A1 = - B1 So you actually think AO AO = -AO AO, and yet AO is not zero. Ditto for A1. <shrug> I have known for a long time that you don't understand the basics of GR, differential geometry, and tensor analysis. I had not realized you don't even understand the basics of arithmetic. Tom Roberts tjroberts@lucent.com >> A_a B_b = -A_b B_a. >> A0 = B0 , A1 = - B1 [quoted text clipped - 3 lines] > differential geometry, and tensor analysis. I had not realized you don't > even understand the basics of arithmetic. I certainly found that little exchange educational. And, thinking back (I'm certainly not looking) at what he's posted previously, I suspect that this 'skew symmetric A_a B_b' is something he uses all the time. It's sad to come across somebody with such severe delusions of competence. [nothing] see my previous posts. Ken In article <dj635h$phg@netnews.net.lucent.com>, Tom Roberts <tjroberts@lucent.com> writes: >Ken S. Tucker wrote: >> A_a B_b = -A_b B_a. [quoted text clipped - 6 lines] >differential geometry, and tensor analysis. I had not realized you don't >even understand the basics of arithmetic. As a rule, cranks are highly consistent in their ignorance, since in order to remain ignorant of one's ignorance, one needs to be uniformly, accross the board, ignorant. Mati Meron | "When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same" Thanks Mati... mme...@cars3.uchicago.edu wrote: > In article <dj635h$phg@netnews.net.lucent.com>, Tom Roberts <tjroberts@lucent.com> writes: > >Ken S. Tucker wrote: [quoted text clipped - 11 lines] > order to remain ignorant of one's ignorance, one needs to be > uniformly, accross the board, ignorant. Mati has that right, Robert's and Low are fuckin with tensors. Tensors form relations, can't sub invariants into components! Way to go Mati! Ken S. Tucker |
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