Removing Redundancy in Algebraic Structure of Relativistic QM

0:00 ...27 on the 5th of August 2005. Unfortunately, the last talk by Grenville Kroll about algorithmic permutation of the Torah was lost from this recording. presumably because the recording button didn't latch down. It looks as though it was in playback mode all the way through the talk. Anyway, the next talk, beginning shortly now, is by Peter Rowlands called Removing Redundancy and Relativistic Quantum Mechanics. These comments are being added by Tony Booth. This is very nice, it's intense, but when people ask every customer, I will give you two. Okay. So please, next speaker. Okay, is everybody ready? It's not Peter. Wait for that to pay. I've been for a long time working on this relativistic quantum mechanics and streamlining it all the time and to me it's so streamlined now that it's easier to use than non-relativistic quantum mechanics now one of the things I've been trying to do is find out why what's the fundamental reason why it seems to to do this. And it appears to me that it's getting rid of a lot of redundancy. And so, hang on, that's it. Yeah. And I think this is what's happening. It's all about coordinate systems and that kind of thing. Obviously we do a lot of transformation of coordinate systems. But when we do a transformation, we're not necessarily transforming like into like. For example, if we use polar coordinates, we go to the privilege of a particular point origin of the radial coordinate. So we're privileging a point, and that's great if we want to do that, where we've got singularity, black hole, gravitating planet, that kind of thing, point charge. But suppose there isn't any singularity there. If there's no singularity there, what we effectively do is we're sort of introducing an artificial one. So we've got to be careful when we transform coordinates. And I think this is what we're doing here in a way, when we do the conventional version of quantum mechanics.

2:30 So, if we use matrix, I've recently reviewed a book by Jack Cuypers on quaternions, which I thought was a very good book, and he was showing about great circle navigation, and he showed that if you use matrix method to model it, by the O3 group, you've got a singularity, rather than if you use quaternions, which are a natural three-dimensional object. so it's not only an inconvenience obviously singularity is a mathematical inconvenience and we know about these things because we deal with them in that way but I think also they cause a problem because if you've got singularity it acts as a barrier and if it acts as a barrier, first of all you can't cross the barrier and secondly you might keep trying to cross it. It's a bit like a gramophone record when you get stuck it goes back to the beginning and tries again so in my view it builds up a lot of redundancy in the system and I think this is the way we do the Dirac equation in the conventional way. So if we look at the Dirac equation, this is the conventional version. Well, it's, in fact, conventional or otherwise, it's still the Dirac equation, whichever method you use. Because people talk about the gamma matrices, but, of course, that's not the case at all. What these are is gamma operators, and the matrix version is only one way of interpreting them. We don't have to do it that way at all. So I'm still going to use the gamma operators, but I'm not going to use gamma matrices. I'm not going to represent them by matrices that's correct, I'm not going to use matrices but I'm going to show why I'm not going to use them by matrices if we use the conventional equation it's not actually all that symmetric Dirac improved the symmetry of quantum mechanics but it's not entirely symmetric because this thing is a 4x4 matrix, this operator and this wave function is a four component spinner and each of the gamma terms of course is a 4 by 4 matrix so it's not actually symmetric as much as it could be and also it's apparently compact you can get anything much more compact than that in fact it's the only equation in Western Australia

5:00 even in a more compact form than that so you can't get a much more compact form than that but in fact it does in fact contain a lot of redundancy and I think what happens is if we use a matrix representation create a kind of fault line. It's a bit like having a geological fault. And the big problem is the momentum operators, the d by dx, d by dy, d by dz operators. That's the problem. They're not rotation symmetric, but obviously physical momentum is. And I think what we're doing is representing a three-dimensional quantity by a two-dimensional construct. And this is It's a million-dollar question, by the way, because I think with the conventional version of represented by matrices, we cannot actually get a wave function that's any good for the proton. Because we cannot do that, we can't understand why the proton has mass. If you can explain why the proton has mass, there's a million-dollar prize waiting. So it is a million-dollar question. I'll do that tomorrow. Okay, maybe you will. Yeah. Could I ask Dr. Ryan, Peter? I want to ask you in what sense the momentum Well I'll show you, I'll come to that Well can you wait for me to come to it then I'll show you If we look at this here we have this is where it originates in my view this is where the thing originates we have quaternions, square root of minus one with that multiplication rule I've done them in red delivery because I'm going to use them in red throughout now if you complexify quaternions by putting an ordinary scalar eye in front we get the square root of 1. And if we multiply complexified quaternum by another, we don't actually get the third complexified quaternum. We get I times that quaternum, complexified quaternum. So we get IJ IJ equals I IK. We can't get rid of one of the I's. isomorphic to that algebra are two other algebras. One is multivaric vectors, as represented by Hesterness from 1966, numbers, which, in fact, are the same algebra, but instead of representing I, Quartanian I, we represent vector I. And these are multivaric vectors. And we notice that these multivaric vectors are J equals little i, ordinary complex I times K. And that's very important, because if we take the parallel Pauling matrices, we get the same thing. Sigma X, sigma Y equals

7:30 sigma i z. These are actually the same isomorphic to Pali matrices. But it's important. The Pali matrices are not quite isomorphic to the gammas. Oh, no, no, not at all. But I'll show how they're connected. Will we know how they're connected because they are? They are, yeah. It's all Clifford Algebras. So these are a conventional representation of the gamma matrices looks like this. Now, already we see they look different. These are a conventional representation. One of them has got complex numbers in it. The other two have got real numbers. And also, one of them is skewed compared to the other two, this one at the end, sigma z. But this is part of representation mechanics. Of course it is. I mean, you can multiply those things by e to the r. But I'm saying it's not a good representation. And I say it's not a good representation two-dimensional object, complex numbers representing a three-dimensional object. And I'll show you why it's not a good representation when we extend it further. It's an almost entirely arbitrary one. It is. It is. But it's not a good one. It's not an ideal one. It's got faults with it. There are only two degrees of freedom which are the complex plane. We don't have three degrees of freedom as we have with quaternions, for example. And this is, I think, what Cuypers was really getting at when he's talking about O3. I don't think he necessarily knows that's the reason, but he does know it's a problem. So one of the three matrices is defined only as the product of the other two, and that's why it has to be complex, because it wouldn't exist if we didn't define it that way. And also it's asymmetric, skew-symmetric. So that's the problem with it. Now, of course, the gamma matrices are constructed from these things, and so they inherit the problem. So these are the typical gamma matrices. These three on the right are the gamma matrices made up out of the sigma matrices and that's gamma naught and gamma 5. So this is what we get for the gamma matrices. So if we do gamma 1, gamma 2, gamma 3, you see two of these this time are complex and one's real. So they're not the same sort of animal. And I say, this is what I'm beginning to say why they're not rotation symmetric. You can actually rotate them, but they're not the same sort of thing. They're not equivalent. the x-dimension is equivalent to the y-dimension

10:00 and so forth. And we need the other two as well. And if we express the Dirac equation, add all the turns together, this is what we get. We get this. We get d by dx plus i d by dy. Already we're losing symmetry between x's and y's. And we also mix up our energy term, which is d by dt with the mass term and when we do the free particle amplitudes this is what we get we get an object like that px plus ipy which doesn't exist in nature we don't have momentum coordinates in that way we don't do it and we get pz separately from them but we get px plus ipy and we also get e plus m so this is just a mathematical I know but it's not a good one you why that's not a good one because we cannot actually represent px py pz on the same footing to represent the protein i say we do need to do that there might be a reason for that and that is that the particle has a spin axis yeah but i don't think that's the reason but i don't think it is i'll carry on i don't think that is the explanation at all the solutions also have different phases we've got to have two phases the top two of those solutions have got to have plus e and the bottom two minus e so we've got to always split our equation into two we've always got to split between a plus e and a minus e we can't actually join it all up so we've always got to have two positive and two energy states and we've inextricably mixed e and n and as I say these terms don't have any counterpart in nature So, what we're actually doing, this is well known what we do, but it's considered perfectly okay, but I don't believe it is. We're changing the momentum energy or space-time signature. We're changing it from plus, plus, plus, minus, to plus, plus, minus, minus. By creating one of those as a complex number, we're making it time-like. So, by making one of our momentum components complex, we're making it time-like. So, we change our space-time signature. And I don't believe that we should do. Sorry, come back up. Well, if we do that, it doesn't square with classical... Nor is it good for quantum ideas, in my opinion.

12:30 You've got to get a streamlined equation. And I'll show you what goes wrong. So it makes it impossible... I can't put a direct in this. That's in terms of signatures in the predicted object. Well, that's what it is. This is signatures. Are you changing signatures? I'm changing... They're changing signatures, not me. I'm sticking with the original signature, plus, plus, plus, minus. Different algebra, it could be equivalent, but some of them with different signatures are equivalent. So you could say that, you could say exactly the same thing in the language of Clifford algebra. Yeah. Well, I'm going to use Clifford algebra anyway, in effect. I thought I won't call it that. And I say that one of the big problems is that we cannot solve this million dollar question because we cannot actually get a satisfactory model for barionic structure. And I'll explain why when I come to it. And this is because the SU3-strong interaction, I think, derives from the 3D symmetry of the momentum operator, which I hope to show. And if you don't have that, then you can't actually represent it. The model says you can't do it. So, to summarise the problems of the gamma matrices, not the gamma operators, but the gamma matrices, equation mixed energy momentum and mass terms they take up too much logical space you've got to use 16 objects when you only want to do one operation they lack symmetry there are five terms in the equation only four have a gamma matrix there's a fifth matrix in the algebra and there's five terms in the equation so we've got lost symmetry again and also they defragment they separate the energy momentum terms from each other they split up in those solutions and in the and I think I don't want to see that. I think this is what we've got. He took out of a hat, basically. You've defragmented your disk and you get all that sort of display, haven't you? You get that sort of display. I'm going to defragment that. That's our Dirac equation. Remember that operates with everything split up and mixed up all over the place. That's what we've got. We don't want that. We want this. that's what we want it has to be defragmated and what we want is a set are you going to explain this to you one day defragmating disk yeah, you have to show you

15:00 that's a relatively routine result that the signature is plus, plus, minus, minus in the conventional representation Okay, I got an impression of the representation if it was R. Is it that you're basically taking out the I out of the Y? No, I'm going to defragment the equation. Yes, I'm basically doing that because I don't use complex numbers any longer. I don't use the complex numbers because I use quaternions. So I don't need a two-dimensional complex number. What I need is three-dimensional quaternions, which are then symmetric. so I am going to get rid of that R so we need separate fins for energy, momentum and mass that's what we need so I colour them in these three colours so this is why we defragment because we've created a problem that we needn't have had in my opinion so if we can get rid of that problem then we can solve it so as I say we don't need to and this is well known we don't need to use matrices that was just one representation but we don't need to use it at all. All we need to do is obey those rules. The commutation rules and so forth for these operators. That's what we need to do. So we can simply get rid of those, replace them with algebraic operators. And that will also solve the logical space because we won't need that big 16-term matrix. And I say we need to use an operator system which will preserve the true three-dimensional matrix of momentum And we need, actually, to have a separate three-dimensional system for separating out E, P, and F. And this turns out to be a pattern which repeats itself many times. So we simply are going to combine these two algebras. Very much that, yeah. We're going to combine these two algebras, quaternions and multivariate vectors. Of course, these are only complex quaternions. But this system needs to be independent from this. algebra. And I actually really, what I really truly believe is these two and a half quaternian algebras. You're doing it in a more fundamental way, it's two and a half quaternian algebras. But that's, this pseudo-scaler I don't believe is really what it really is. I think it's half of the... Well, you're an unfinished quaternian system. I don't really believe there's such

17:30 a thing as a pseudo-scaler. It doesn't really matter, you don't need to know that for this. And I think it's incredibly less complicated than it looks. We soon get rid of the vectors, in fact. So, just to remind anybody who isn't very familiar with multivariate vectors, what's the difference between them and an ordinary vector? If simply that you'd define a full product, AB equals A dot B plus I A cross B, which is Y, I times I equals 1, and so forth, and I times J equals I K. The full product vector itself is just A squared. Well, Hesterner did great work in this, and he showed that a multivariate vector P or Dell, even in the Schroding equation, could lead to the automatic generation of that extra term which creates the spin. This is one of his main results. Whose word is this? Hestoners. Yeah, 1966. He's still alive, he's still working. Space-time algebra. Very good. You'd be very impressed with it. So we can choose any mapping we like. These are the mappings I've chosen. So I choose to represent my gammas that way, If you also choose to represent them that way, you'll find they're exactly equivalent. If you multiply those out, you get the same commutational relations. But I need two representations because I need to transform from one to the other during the operation. So I need two representations. But there's a very large number of other representations, but they're all equivalent. If I choose the first mapping, I can't actually use my colour coding here because this doesn't allow it. you know, if you transfer from the equation editor, it doesn't allow you, not on the way I do it. So I can't use my colour coding. But this is now the Dirac equation as I would write it. And all we need to do now is multiply from the left by this is using representation. A, I'm going to now, this is a very trivial step, but it has profound consequences. Because if I do that, I suddenly get this. Which switches the mapping to B, but at the same time incorporates gamma 5. into the equation. So now we've got perfect symmetry between the five terms. Gamma 5 into the equation. But you see it's exactly equivalent, except I multiplied it on the left by j. And what effect does that have? Well, it has a very profound effect, because now I've got my three bins. I've got my bin for energy, the d by dt term. I've got my bin for momentum, with all the terms exactly symmetric. And I've got my bin for mass.

20:00 so this is like what we do when we use complex numbers for many things in physics we put one thing in one category and another in another we don't need to mix them there I didn't actually do it that way to start with I did it from this, which is a completely different concept this is the way I've always looked on physics I just thought I'd mention this at this point because it might be relevant to what I'm going to say next what I did was start with this symmetry between these objects and stick the charge operators onto those and so I then used that to create these energy, momentum and mass terms from time, space and mass and that's very old in the work I've done on that but I then realised I could then use it in this other way and my other argument is that one not only did that one not only changed the black terms combining them with the red plans, one also changed the red terms, so one split the symmetry between the charges as well. So one became pseudo-scale, and I said that was the weak one. One became vector, which was the strong one, and one stayed scale, which was the electric one. And that was my argument on that, and I've done a lot of work on that. So now we have a combined formula. I'm now going to combine my momentum terms, because I don't need any longer I don't know whether I've actually done something I've wanted to do there can I go back a bit what I really need to do is to say that if one solves using say a plane wave or something like that then one gets this as a instead of d by dt you get e, p and n unfortunately let that slide out If I solve, if I put a plane wave solution in, if I put a plane wave solution into the original equation that I did using these objects, I would get that that was the amplitude. Maybe I'll say that later. But the important thing about this, and this is, you know, once I had a disagreement with

22:30 Basil Hale over, was that this meant that the amplitude of the wave function for a plane wave was a nil potent, square at zero. Because once you've gone that, you've squared it, you've got e squared minus p squared minus m squared, which of course is zero. I think I've got that wrong as well. It should be e squared minus p squared plus m squared. these should be minus sorry I've got that yeah I'm sorry if I put the plane wave solution let me go back to the equation into that equation if I put a free particle into that and that's my solution put that into that I will get the amplitude A will look like the amplitude A will look like that You know, once I've done my differentiation, I'll get that. So if I put that into this equation here, if I put that, d by dt, I'll get an e there. If I put d by dx, I'll get a px and a py and a pz, which will add up to p, and then my n will stay the same. For any, that's a free particle, but I'm going to argue that that's not the only solution, that's no potent. I think they all are. That's the next step is to say, that should be e squared minus p squared minus n squared, so that's square of zero. Have you got any? I still don't understand what you're saying about that. You set up the operator on the left-hand side of the... Yeah, and then I'm putting this solution in. This is a free particle. If I differentiate that respect to T, I'll get an E term here, which will go on the K. Sure, so the operator does what it's supposed to do. But the A can be anything. Well, the A can be a scalar times that. Yeah. It can be any scalar times that. It cancels through. Yeah, but I'm not worried about a scalar term. Yeah, it can be any scalar times that. will be the key part of the amplitude. The amplitude is the absolute square of A. Yeah, but the amplitude is zero, not the square is zero. But A is the scalar, it's not the non-commuting guy, and... No, it isn't, not in the Dirac equation, it isn't.

25:00 The amplitude of the Dirac equation isn't the scalar. No, okay. But I'm still confused about your discussion of the nature of it, Because you're saying that the A is operators. Yeah, well, they are. They are in the direct equation. No, they're not operators. They're not operators. They're algebraic terms. Oh, I'm sorry. I just understand. A thing people ought to realise about that particular form is that it doesn't normalise. In other words, you can't treat phi as a probability density. Not with that particular work equation, because it's absolutely flat and it assumes that the volume of the space is infinite. But in fact, I'm going to argue that all of them, all amplitudes on this, are in fact nil-potents. Well, you're generalizing the notion of amplitude, so... Well, okay, I may be using the wrong term. Because amplitude is supposed to be... You're usually thinking of amplitude representing the probability. Yeah, but I can actually do a probability density quite easily. So if I write this thing, it really should be a four-component vector or spinner, if we're one's doing the Dirac equation properly. So that should be a four-component spinner. And in that, you've got fermion and anti-fermion, spin up and spin down. identify here. If I need my energy to be... If I need fermin-antifirmin, that's plus or minus V. If I need spin, it's plus or minus sigma dot P. And I'll show you if you can actually use P instead of sigma dot P. Have you got any feeling for what it is that Basel doesn't fly? No, I do. Yeah, what I'm saying now is E and P can be either operators or eigenvalues. It's the eigenvalues the word I should have used. So they can be either operators or eigenvalues. These E's and P's don't need to be, you can consider that you consider that as the operator, E as the operator, in the normal way we write the E operator. And you consider P as the operator, in the normal way we write the P operator, and for a free particle they'll look like that. Yes, I mean they're computing operators with a continuous spectrum and therefore

27:30 you can write down the scalar equation in terms of the spectrum yeah but i'm saying what what in this the operator and the eigenvolume look the same yes yes well they always do when you have a continuous spectrum yeah okay uh but the interesting bit about quantum mechanics is when you don't have a continuous yeah but but this can do anything actually as it happens so really the only independent information in that is the first term the others are So you automatically change your d by dt or your i. You go through an automatic process of change. So once you've established the first term, that's the only information you need. Now, what you can do is, in fact, this thing here, now this is a slightly... This operator here totally determines the nature of the wave function term. It totally determines it. If it's nil potent, it totally determines it. So if that is the nature of the operator, then I don't actually need to write a wave function down at all because I know that that wave function must be such as to make the eigenvalue nil potent. an eigenvalue can't be nilpotent an operator can when I say nilpotent, I mean a square of zero it must square to zero well what you're saying is that the coefficient a is now interpreted as a vector if I understand the same thing but that vector can be thought of as written in the fraternians and then you can ask what algebraic property does that vector have in the fraternians Oh, well, when I square it, it's usually zero. And so it's analogous to an amplitude, but it's an algebraic in this non-fueled algebra and it's square zero. So it's a general one. It's what I mean by, I did put inverted commas amplitude because I knew that there was a dispute about the word. And people don't usually think of those vectors as parts of the algebra. No, but I'm changing things so they are, in effect. So that's why, I mean, it's kind of entitled to use it. this must create its own phase term such that that solution will always be square to zero so I don't actually need to write any wave function with that I only need to write the operator and everything else follows but that's if you're assuming that the wave function is a plane

30:00 yeah, but I don't just assume that I say that's true irrespective because this time I say that one can generalize that principle and show that it works in many cases in all the cases I've done it works that one assumes that this if one writes E as an operator P as an operator that those could contain any field terms you want that they could be covariate derivatives and the same principle would still apply the solution would have to be nil potent but all this stuff is well known that's not well known at all it's not well known to anybody that I know of I don't know where I picked it up but because I've talked about it many times well if you say If you say PQ minus QP equals I, for example, you can stick any function of the coordinate with the P. You're absolutely correct. But I'm saying that I actually know that it must be this nilpotent expression. It must be this zero. And that's different. That isn't spoken about anywhere else. I'm saying I know actually of a thing that must always be you're saying if you know it's always a certain thing you can always do that, I agree with you there, but I'm saying this isn't known, that it must always be square to zero well yes, the point I was going to make is that the functions you do add on to the d by d representation in the p's depend range your coordinates go over. For example, if I have x, y, and z, I just say I don't add anything. It's dx, it's ih, or minus ih times d by dx, and so on. But if I halve the range, coming from north to infinity, say, instead of from minus infinity to plus infinity, I find I only get self-rejoint operators out of each if I add a 1 over r and so on. So you are compelled to add certain things. Okay. Well, I say you are compelled to add certain things to make it no potent. And you're always compelled to have a Koulomb turn. You're compelled to have a Koulomb turn if you've got spherical symmetry. Always. Yes. Okay, good.

32:30 Yeah, so I'm saying you can do this for anything. that to them into seven lines. It's the full solution. And also, these things seem to have multiple meanings. If you fiddle about with it, you've finally got multiple meanings. So I'd say the primary meaning for me was always charge. But if you pre-multiply the nilpotent, you get a vacuum operator. If you post-multiply, I'm sorry, you pre-multiply a nilpotent by that, and you post-multiply the ordinary bracketed term by that infinitely, you'll still get same answer, give or take scalar terms, give or take, scalar values multiplied by. Well, you can take this object here, that thing there, put it in front of that, and then multiply it by that, and you'll still get basically that multiplied by scalar, which you can delete, you can get rid of the scalar. Then you multiply it again by this term, and you still get the same expression, still get a scale up. I think I might have a slide that shows that. Can I just leave that? I may have, I'm not sure. You're the slide that does the hydrogen atom in the second line. I have it, but can I come back? Can I do that later? You can have it. Yeah. Pre and post multiply, no potent, transforms it. So if I do before and after, it does a T transformation if it's K. It does a parity transformation if it's I, and if it's minus J times J, or IJ times IJ, you get the other one, the charge conjugation. So, the lead term, see, this is what I'm now going to do, an interpretation of what those four terms in the Dirac equation are. the four terms are the lead term is whatever the thing is fermion, anti-fermion fermion spin-off, fermion spin-down anti-fermion spin-off, anti-fermion spin-down and the other three terms are all its vacuum, as I would put it reflections through the strong interaction, the weak interaction of electric so the kind of vacuum terms that go with the lead term the lead term determines what it is but they all have the same pattern So this is my interpretation of how the charge operates. It's a kind of discrete partitioning of a continuous vacuum

35:00 responsible for zero-point energy. So you partition it according to the weak interaction, either electric or strong. And they're completely independent of each other. Right, yeah, that's showing the CPT symmetry argument using this. So I multiply by I front and I behind. Now I am confused because it seems to me that you chose a very specific co-ordinatisation, which is very beautiful, and then that's going to change if you co-ordinatise in some way. Why should individual parts of those co-ordinates correspond to fundamental things? Why should they? Because I think they do in the initial instance. That's where we get that combination of E, P and M from. I think it comes from that originally. Torrin here. I mean, I'm saying if I start with those, I get these answers. Okay, carry on anyway. Now, suppose we've got an interaction vertex, what we're talking about. So, the size of the E and P terms, if we've got an interaction vertex, will determine whatever balsonic states created. So, there are three operators involved, so there's three balsonic states. and we'll explain what those are so that's an interaction vertex there and all the interaction vertices in this are either 1 or 0 basically 1 if you normalise it or 0, nothing else they're all scalar all the interaction vertices that you get they're all scalar so if I did this it doesn't look like it here if you multiply this out you won't get a scalar but you've got to remember there are four terms I'm just writing the first term only for convenience minus terms going down there. And when you do the full scale of product, the full product, that's what you get. Now, if the E, P, and N values become numerically equal, then we get an actual new combined bosonic state. Those are the kind of states we get. If we do that, remember, of course, that these, you've got four terms in each of these. I'm only writing the first one because all the others are automatic changes in sign. If you If you do that one, you get spin-1 bolson.

37:30 Do that when you get spin-0 bolson. This is how I interpret a bolson-stein condensate or ferry-phase or any of that sort of thing. It's something of that nature, it's fermion on fermion. Think of, I don't know that right. Yes, that's all right, that's the spin-0 version. You can do a spin-1 version. Interesting, the spin-0 bolsonic state can't be massless. find that out and getting an answer, you get zero. You cannot have a massless spin bolson state in this formula. This is what the Higgs mechanism tells us. We cannot have a massless spin zero bolson state, a goldstone bolson don't exist. But they don't exist in the algebra here. Well, yes, this is what you asked me earlier, and this is what I meant. If I keep multiplying that forever and ever and ever and ever by that, I don't get any change from that, other than the scale of multiple. Of course you don't, because they square to a scalar. Yeah. That's right. No, I get this. This multiplying that by that gives me that, always. This multiplied by that always gives me that again, and so forth. That multiplied by that always gives me that again. Yes, but this is a property of, for example, the gammas in the original Dirac. Well, I've never seen this. I've never seen a vacuum term represented that way, and it couldn't be, because there isn't a unified expression for all the solutions together. There's no such thing in the conventional version. So you can't actually write this down in the first place. You can't actually write this term. No, but that's an algebraic property. I mean, you may... Well, sure, of course it's an algebraic property. But it's an algebraic property that doesn't come out in any other version that I've ever seen. and the other aspect of this I don't think you have because you haven't seen that and you couldn't have seen that because there is no other solution which provides a unified single state it's always four different ones that is a single one a single one there's no other information you need I think you'll find that's true of course it's algebraic it's purely algebraic, that's my argument that if one can do it purely algebraically then one doesn't need to find physical explanations for these things. It's all part of the algebra. And if one does this,

40:00 notice that if I multiply that by k on either side, that would change. Well, that one would stay the same. That would change. So I'd get alternate anti-fermions there. So in other words, I'd get a bosonic state created. A fermion would create a boson and a fermion would create a boson. Now what's that telling us? That is telling us that in this formalism each fermion also creates its own boson vacuum states. So it's supersymmetric with its own self as a boson. Each fermion is supersymmetric with itself as a boson. And each boson is supersymmetric with itself as a fermion. And I don't know any other algebra that does that. And it's true whatever kind of boson state one has. And all these will be true at once. right, let me carry on so none of those boson states have any effect on the ferment each is only supersymmetric and so, this actually removes renormalisation necessity in the case of a free particle, and I can show that mathematically, while renormalisation of interactive particles becomes rescaling you haven't determined the charge value concept from what it normally is. You still do it, and I've done it, of course. This is a free particle. I can't give the whole calculation, but if you actually do a perturbation expansion for a sort of coupling of virtual fault and etc., you get something like that, and when you work that out, it's zero in the nil-potent algebra. So the psi-1 state for this is actually algebraically zero. So the free particle doesn't need any renormalization, and I say that's because one has removed that redundancy. That barrier is gone, and therefore that complete change of going back to the beginning, starting again, isn't happening. Also, you don't need to worry about the infrared divergences propagators. You don't get that at all. So this is the conventional, I think you can only just about see that's P slash there, P slash. You see it's P slash. And you get a singularity or pole where you get that, of course, in the conventional version. done. And they, of course, do a contour integral, a messy thing, but eventually they get this as a result of their contour integral. But notice it's got backward time, backward time, positive energy, negative energy. Two separate halves, split into two, because of this two-two

42:30 space-time signature that they have. However, nil-potent formalism is totally different. if one writes as propagation nil-pultric formalism that is does not that multiplied by complex conjugate does not does not become zero it becomes that it becomes a positive scalar on the bottom so this does not have any pole at all there's no infrared divergence whatsoever and the reason is of course you're packaging all your four terms at once you're not splitting into positive and negative energy there's no barrier between which you've got to cross between positive energy and negative energy that's one of the things Basil didn't like, because he values that because it shows him where he gets his complex numbers from. He does something similar, but I don't need it. But, you know, his method is aimed at something different. He wants to show two versions of quantum mechanics the same. I just want one version of quantum mechanics which I've been used. I mean, it's a totally different reason for doing it. I think that's one of the things he didn't like. And this is what the integral now becomes. You see, just one term. These column vectors and row vectors, these things, with all these things in, and you multiply them out, you get a single package, one integral, the lot. No messing about at all. If you want, that's, I was doing fermions there, but if we can do bosons here, you can do three different boson propagators with three different types of bosons. Same thing again. Oh yes, that's what I explained earlier that's the thing I explained earlier about the meaning of the four terms in the Dirac four-spinner What is here, what I've really got is my differential operator my operator is a four-spinner like the weight function so I don't need the weight function because the four-spinner determines everything now, can we actually do anything with baryons, protons so, we've got three fermionic components to which we assign colour, this is the conventional version of course I don't doubt that for a minute but, can we have that no, we can't have that because it's zero, so we can't do that but why three why don't you use the fact that the momentum is a vector operator if we did that

45:00 momentum in successively that term, or that term, or that term I could get those, and I would then get these just multiplied by a scalar notice it's very significant if there's a sign change in that one if the one in the middle changes sign but if you can, if you believe in both signs, if you have both signs of the plus and minus, then then you can do it and let me just show you how I would do that there we are state vector like that privileging so you start saying PY and PZ are zero and you've got PX and it can be plus and it can be minus so that's one possible state or that can be zero and it can be PY and so forth but only one of them can be at one time so you've got six allowed phases gauge invariant of course obviously with an automatic mechanism for switching which of course they don't do they're there all the time so we get something like that if one ignores the black p's zero those the black p's are zero only the red p's are real only the red p's are actually acting so we've got six possible phases for which we could get a non-zero solution exactly the same symmetry as the strong sorry let me go back a bit just a sort of pseudo animation here just imagine the red thing being what's carrying it. And now let me put that next to that. And I've now got the same symmetry as we use for the conventional version of the strong interaction. But you can only do this if your three momentum components are on the same plotting. They're not on the same plotting, no chance. So, yeah, so it's got the same structure as that, which means it has an SU-3 structure. That is an SU-3 structure, SU-3 structure, eight generators, and so forth. I'll leave that out for the moment I'm just arguing that we must have a strong interaction as we now know it with a linear force because we must have a constant rate of change of momentum which is non-local so it doesn't depend on the separation of the components whatever that may be and of course we couldn't do this without having equally plus and minus for the p operators what does that mean if we got plus and minus operators, it means that left-handed and right-handed must be present at the same time. You've got

47:30 the same sign of energy, and you've got both P and minus P, at the same time you've got both-handedness. You've got both-handedness is you've got mass. I'm arguing that one can actually explain this problem. And it's a Higgs mechanism way of getting mass as well. so I'm arguing also that the baryonic representation can only exist as a unified entangled state it's not actually three independent terms, it's no more independent those components can be no more independent the components of momentum or space can be independent of each other and this of course is what you'd expect this here couldn't be done if you got px plus ip, why you can't do it, because they're not on the same actually build that structure at all and because you've got a fine nature of any vector exactly the same thing will happen with the structure of the proton as is found in our experiment and one can also structure the ones, of course remember these are all there's almost four terms in these things I only write the first one because that's all we need so we'll have six pulsons of that form and two combinations of those, of course the reason why we don't have the third one is because that's not independent information. And again, these things cannot be applied to those other things unless they actually reverse the sign in the operation. So if I apply this thing to change an X to a Y then I've got to change the sign of the thing in the original baryon to do it the original P. So even the actual act of applying these things means I have to change the sign in the act of doing the interaction which means I've got to have both plus and minus signs. Therefore I've got to have maths. and since we're talking about H mechanism let's have a look at this so if we imagine a virtual phimonic state no massive vacuum let's imagine something like that if you've got an ideal vacuum you've got exact symmetries and so let's transform it to C we get that and that of course is indistinguishable from that not an indistinguishable but actually indistinguishable so we don't need a balsonic state for that transformation let's suppose our vacuum is degenerate in some way and it's degenerate under sea transformation, then we will of course need such a state but we cannot have that as a balsonic state, we must have

50:00 that as a balsonic state because that can't exist that water must be zero, so we must have mass in it, so that's a nice simple explanation of the Higgs mechanism I thought the Higgs mechanism was degenerating there, giving an idea of what we need to talk about that is exactly what it is It's having two states of... It's... The reason why we can't have that is because we can't have fermion and anti-fermion with the same state of spin. That's effective. We can't have left-handed fermion and right-handed anti-fermion unless they've got mass. That's effectively what we mean by the Higgs mechanism. And that's effectively the same thing as saying that this thing here, algebraically, is zero. It's basically the same argument. So, there's a lot more, but I don't think I'll have time to go over everything, but I'll just quickly mention a few other things. I think a weak iso-spin transition is something of that nature. Something with one spin state transforms into a superposition of two spin states. Of course, this may have two spin states to start with, but one of the parts that isn't two spin states transforms into two spin states. and so if we have a look at this we're introducing, this may have no right handedness in it because it's only got one spin state but it's got two spin states so you must introduce right handers, therefore you must introduce mass I think I'll just go over because otherwise it will take too much transition from left-handed to right only a few minutes, because of vacuum let's skip that yeah, I'm just saying something like this is happening and this is using the C, P's and T's and stuff, we get something like that happening, pure transition from left-handed to right-handed electro-weat combination go into all that in detail I could actually show, I think I've got the, what you asked me for, the hydrogen atom, if you're interested in that. Let me ask another question. Yeah. So you're showing us that by working in some plexi-tripraternian algebra with appropriate labels,

52:30 you can see all sorts of different particle interactions. Yeah. Yeah. You can actually represent the interactions that way. Direct equation began. It's all happening in that algebra. It isn't, there isn't even, I mean, these are, this is cataloging patterns, so this isn't, one didn't even need to know about the existence of three-dimensional space for those patterns. No, no. The patterns are all just algebraic patterns. You do for the post. So how do you know, how do you know what you identify with what? Which part does that? Without prior knowledge of... Well, it's arbitrary, which you call E and minus P, but you just label one of them, it doesn't matter which you are. physics, how do you know what to identify this? How do I know what to identify? How do you know how to read this without knowing the physics beforehand? Well, you don't actually know that this particle is represented by that operator. You don't know that. What you say, these two particles are represented by these two operators. One of them must be one, and one of them must be the other. I mean, if I blindly reached into the algebra and grabbed two things, and I multiplied them, I could find out something about or how they interact in your model, whatever they were. But then you do much more than that. You say this corresponds to that physics and this corresponds to that... Well, I haven't actually mentioned any real particles except bosons. I mean, you can actually do the standard quantum mechanics, by the way, and get spin-hardt and all that sort of stuff. I've done all that. You can do all the standard things. It's not that you can't do the standard things. You can. It's just that I'm doing things that you can't do in the standard things. quantum mechanics with it. I can actually use it as Dirac equation if I want. But I'm trying to get some extra information that isn't in the standard formalism. But I'm not rejecting the standard formalism and just going beyond that. So anything you can do in that, you can do in this. I think I can answer your question, but it's a difficult thing to answer. Do you mind if I just leave that for the moment? I think I'll skip these. There's a lot on this, but I'll leave this, because I'm saying that one can actually do these. Well, actually, one can actually generate these from the Dirac equation. One can see there's only three nil-pultent solutions. One of them is Coulomb solution. One of them is Coulomb's strong interaction solution. And the third one is a harmonic oscillator, which I associate with a weak interaction.

55:00 Well, there's only three that one can actually do mathematically solve the equation in this form and find there's only three nil-poltrum solutions. There's only three potential structures which work to give nil-poltrum solutions. But I'm going to leave that. Because that's too deep. Yeah, you can do all this sort of thing, but I'm going to leave all that. Because that will take a long while. represent by potentials. I'll leave all that. I'm saying that there's only three solutions that we can get. I might just mention how you actually couple the coupling acquisition of mass through the Higgs pulsant. So, I mean, what you're doing is, in fact, you're doing the Higgs as a vertex involving the fermion. There's the non-massive fermion and the massive fermion and that's how it operates. It's quite very simple indeed. I'm going to leave it though. I'm going to give a prediction here though, on the very phase. We've just got the time, haven't we? I'm still very worried about the civic program because as I am the service as far as I do, The main notion was to inject the concept of mass into the master's theorem. Yeah, but I don't do that. You speak a bit about... He doesn't inject the concept of mass. He says we start with a degenerate vacuum and then we get mass. I start with a degenerate vacuum and get mass, doing the same thing. He doesn't actually inject mass. He starts with a degenerate vacuum and says a degenerate vacuum will produce mass. But it's a very complicated way of doing it. Yes, I know, but you've had mass in your equations the whole time. No, I haven't. But I did that, I did. But the Durant experiment has mass. I know. But to do right, because I assumed for a minute there wasn't any mass. Well, I didn't miss that one. All right, well, I started off with a massless term and, you know, got the mass afterwards from the algebra. Assuming, assuming a degenerate vacuum. But one thing is assuming a degenerate vacuum, because a degenerate vacuum is forced upon you when you actually solve the equation. but can I leave that because I'm running out of time otherwise I want to make a prediction I think the pseudoscalar weak charge

57:30 is a dipole, it's a dipole with its vacuum partner and this is what half spin really means I think it's, leave all that out that's not the prediction I wanted to make a prediction in the very phase the spin state is such as would be required for a pure weak transition from that, or it's inverse. Now because the state is necessarily massive, the spin zero state is necessarily massive, time reversal symmetry must be broken in the weak formation or decay of states involved in the very face. Aronoff-Bomb effect, Yarn-Teller effect, superconducting Cooper pairs, all that sort of stuff. All those things, I think, break time reversal symmetry. Anyway, we'll see. So that's my conclusion. Weak, strong, electric interaction, the acquisition of non-zero-as can all be related to the null-fonte state vector, particularly relative signs associated with that. I didn't get much time to talk about that. I did talk about it a little bit. Let me just see if I can go... You asked me about solving the hydrogen atom. I think I can show you that. What you have to do is write that in polar coordinates, sigma dot del term, in polar coordinates. And this is what I get, writing it in polar coordinates. Now, if I do that, and put that into the Dirac equation, that's just writing the delta in polar coordinates. If I now put in a Coulomb term there, I get... So this is the equation I'm starting from. This is the Coulomb term, this is polar coordinates, and then we just use this standard solution. we assume that and then we get that and then we just get a few simultaneous equations which are easily solved not as quickly as that though just as quickly as that I can show you the calculations they didn't impress me particularly it doesn't impress me, it just shows that this works and it can in fact do the strong interaction and the weak interaction as well what it's actually telling you is that at the origin of coordinates the singularity is genuine oh yes in this case

1:00:00 now there's a corresponding thing in the Schwarzschild solution in general relativity where there seems to be two singularities there's the one where the factor M over C squared over R is 1, and you think, gosh, that's going to blow up nicely. And there is also the one when R is 0, and by making transformations you can show that the R equals 0 one is the only genuine one, which is why physicists say when you go past the spout shield barrier you don't sense anything i'm not actually trying to impress with this because this is standard but it's but the reason why i'm doing it this way is to show that this method can do it and very quickly but it can also solve which things are not standard and i have done that all right which there's no no known solution for this thing here i don't even need to put that in i can just demand spherical symmetry and that must be there if I have a point charge I don't need to put that in that has to be and otherwise I don't get a null-potent solution so I'm not assuming that there's a Coulomb thing at all I find that there must be a Coulomb solution because of this term I can't have a Coulomb solution I can't have spherical symmetry unless I have this term in there I can't have null-potency unless I have that term in there Yes, because Mr. Watson is slightly worrying when you're through. I'm not quite sure of the status of the nil potency when you get to this state. Well, I say it's absolute. In other words, it's the basic axiom of your theory that it must be nil potency. Yeah, well, if one looks at the way the Dirac equation is normally solved for hydrogen atom, what you find is that the eigenvalue term is assumed to be a constant, In this, we don't assume it's a constant, because we don't actually go through that equation. And you do insist on this nil-potency. Yeah. But I'm saying that what the Dirac equation is really doing is doing the nil-potency. But it does a little bit of a fraud because it actually assumes the eigenvalue terms of constant, which, of course, it isn't, because it's a function of R. And this is more technically accurate than that is.

1:02:30 I know that works, to find something that would work, that was more technically accurate, and this is more technically accurate. That only worked because we're not really truly using the Dirac equation. We're using a little bit of a fiddle. But we're not using a fiddle here. And I'm saying, well, what is the Dirac equation doing when it does that? In fact, it's doing this. So if I make that my action, can I solve anything else? And I can solve lots of other things. You mean it's a fiddle because we're using the matrices? No, it's a fiddle because we assume that the eigenvalue term. It's a constant which it is not. So we ignore that when we differentiate. We shouldn't differentiate how long would we ignore it. But if we use nil-potency, we're not actually doing that at all. Never just put in a historical remark. You go back beyond estimates. Well, he's not the originator. He just showed it works in physics. Well, Sure. Well there's a paper by Luntzos in Psychiatry of Physique sometime called The Arash's Electronium Schwerver, which uses britannians to express the electric vision. Now, Eddington was doing this in, of course, no, I'm not saying that they could do all this stuff, but just as re-expressing Dirac's equation by two Quaternian algorithms, Eddington was doing it in a somewhat confused form in 1940. Oh, no, there are Quaternian versions of Dirac, and Conway did one, Arthur Conway. Yes, John Conway. No, Arthur Conway. Yes, A.W. Conway. Yeah, he did one, but this nilpotency is what makes it powerful, in my opinion. Yeah. I have a slide show. I have one for my 1950 PhD piece this week. No, no, you've done work on that. But, you know, whatever anybody... I mean, I think there's no doubt that this Neil Bolton's version is quite original. I've not seen anything like that. Just one slide, I'd like to have. I'd like you to emphasize... Yeah, I have actually referred to some touch papers. The assumption of the old code of 2016 and even 12 and then... Well, I have. Yeah, yeah. No, I'm saying... I'm saying... But what the hell is the Dirac equation doing by assuming that this eigenvalue thing is constant? Well, it isn't. This is what it's doing.

1:05:00 If that operator is genuinely nil potent for one on the left hand side then it doesn't matter what I suppose what you've been saying is it doesn't matter what part side you put there Well you've got no choice, it's simply determined solely by the operator, so all you've got is an operator and all you've got is the first term of operator, because the other three terms are automatic information in a completely streamlined form which makes it very easy to use well when you solve the hydrogen atom normally for example you have to have two sets of equations interacting it's a right bloody mess but but when you've got um just just one set of equations and it's all packaged in one then you don't have to do that interaction it cuts down alone it's a bit there is a corresponding thing that you can do at the newtonian level with the newtonian hamiltonian oh sure and what but but it limits you to saying okay let's have small r and see what happens see one of the things you get you get the one over r coulomb force you can't avoid it right i'm sure i'm sure that's absolutely true and what you can do with uh what it didn't emphasize in Eulerian of the Lagrangeas, I'm not sure which is which, at some point in this coordinates. You transfer the spinner nature from the wave function to the operator. You make the operator a single turn, first of all, get rid of the 16-turn matrix and then you can actually expand it into being a row vector with four turns in. So that becomes the spinner. So the operator becomes the spinner so you don't need a spin away function as well but the operator is only a spinner because all the other terms are automatic if they're not different terms if they're automatic so they're just drums of the first time so only the first time matters there is something that does bother me not about this if you start off with hourly operators you notice I've used the word operating over the matrix. Operators, emissian operators with those processes, you can generate formally quaternions, I am told. Well, yeah, because you must be able to. Hang on a minute.

1:07:30 If you started, therefore, with, I suppose that's what Dirac did when he constructed Yes. So nobody used them. They still are. Well, you may remember a long time ago, I did a long paper about averaging, the nature of the connection between the quadratic Hamiltonians and the linear ones, and I said it was the averaging that happened over measurement, because if you did it over a short time and collected lots of information, you had to use high frequency. you disrupt the whole bloody thing because there's too much energy. So, and to give him a due, Dirac more or less says this in his book, but I carried this out for a little while, and found that the only operator rules that really fitted were the Pauli ones. I couldn't... Yeah, well, I mean, when one solves Dirac normally, one shows that they're the only ones that can work. But all you have to do is have those commutation relations, you don't have to have anything else. You don't need anything else, you don't need to say, oh, there must be matrices. No, no, no. And that's not, you can certainly use matrices if you want, but you only get so far with using them. And I say you can't solve the variant structure, or you can't solve the proton. Well, yes, I mean, not only Dirac produces them out of hats, modern writers produce them out of hats as well, with six and eight and God knows what. They're quite happy to do that. Yes, yes. Anyway, sorry, we need to get on to this second presentation. So I'm starting this, and then Vanessa's taking over, because this is physics and biology. So this is a bit more speculative, in the sense that what we're trying to do here is is to see if there are some sort of fundamental structures that seem to repeat themselves that are to do with replication, reproducing, and all that kind of thing, that somehow can show some similar patterning in physics and biology. Now, at the moment, it's exploratory, and one has to make guesses and see if they work.

1:10:00 So, let's just see what we can do. Sorry, I'll turn that to start. No, we're carrying on. This is the second presentation. Vanessa's taking over halfway through. I'm starting here. Vanessa's finishing. It's a joint presentation. It doesn't say so there, but it is. You don't have to know what the previous talk. No, it helps, but you don't have to know. So, fundamental questions. How do you get asymmetry from symmetry? That sort of thing. Now, there's some tantalising hints. Now, what I hadn't emphasised in the previous talk ideas from, because I don't get my ideas from filling around with a Dirac equation. I get them from fundamental algebras. And I now represent them in this computer model, this rewrite system. And all this three-dimensionality is important to me. And I see three-dimensionality, in fact anti-communicativity, as the complete foundation of discrete numbering. Because to me that's what happens in physics, and it must happen elsewhere. Are you saying that you're investigating the rewrite systems No I didn't. But I did investigate fundamental symmetries before I did that direct stuff. And that direct stuff is quite a lot of my work to start with, but before that I was working on other things that were related. So this rewrite system comes after that, and it starts from total zero state. And I wish I could talk about it, but I can't. I think I talked about it last year. So people have to try and remember what I talked about last year. It's in the proceedings anyway. so all components are necessarily dual or conjugated we are forced into anticommutative operations to avoid defining zero I don't want to discuss that but I want to say that's the basis when I say anticommutativity is to my mind the only true source of discreetness in nature to me the fundamental things that are discreet are anticommutative as well yeah well I don't but if I do the rewrite system I say the other way around I mean one could say that but from my point of view and what I do in this rewrite system eventually and this is, the rewrite system was worked out with my computer colleague Bernard Dyer as I mentioned earlier this generates an infinite succession of identical closed anticommunitative sets which are effectively a Clifford type algebra they've gone on to infinity and in fact just as in Clifford algebra you're basically doubling it but the fundamental reason why you're doing that and I say that the first four stages of this have the same algebras as those operators

1:12:30 those objects there I only meant to take this use of these conventional words on the right-hand side seriously. Yes, totally. I take them seriously. I'm sorry, this is where I have been unable to... No, OK. I'm sorry, but I can't go into that, because otherwise we won't get into the biology. Yeah, it doesn't really matter. I'm looking for patterns. You mean you're not... I don't have to... this doesn't have to be taken seriously. Well, you can just suspend your belief for the moment. suspended belief in thinking that time, that mass and space are the same kinds of things. Well, sorry, that's fundamental to my way of thinking. You might as well say, I've got cows and horses and muchness and identity. No, you can't agree with that, but that's my view, so I can't alter it. But it's not a view, it's a nonsense. I'm sorry. It was pillared by Lewis Carroll. but Lewis Carroll didn't live in the 21st century I'm sorry I can't argue with you on that, we can argue later it's irrelevant anyway can I ask a very brief question, have you ever looked at Neil Nester's approach to political algebra, I really think you should give me the references after this is my sort of way of looking at things, I just think we just duplicate this way And this is why I was talking about half a quaternion set, because, you know, you're not finishing that one. You see, you're kind of finishing that set, I1, with J1. You're finishing I2 with J2. And I3, by the time you get to order 64, which is Dirac algebra, which can be made up out of these, but then you're not finishing that set. And that's really where you get your scalar. These classes are essentially tensor products. Yeah, well, they're just multiplication. Yeah, just ordinary product. just ordinary product of if I multiply one by J1 I'm just multiplying everything in those brackets that's what we are a tensor product it is effectively a tensor product but it's not I'm just that's why the order goes up by a factor of 2 and that's what you get you know that sort of stuff so you don't need to go into all that and this is where I've always come from I have to explain that on a separate occasion because I couldn't fit that in and talk about this biology as well.

1:15:00 Can I talk about that with you afterwards? It means something to me though. I talked about it last year. It's in last year's proceedings. I think you'd be quite interested in it. What it is, is that, you know, you can write a, you can express a lot of computing in terms of a rewrite system, I think that creates its own, or it starts with a set of rules, an alphabet, and that alphabet then structures whatever happens to the system, and it responds accordingly. I understand vaguely what this meant by the phrase, what's on a rewrite system, but I don't understand the word. No, I did work with a man called Bernard Dyers on this, a computing colleague. And what we did was generate this mathematical algebra from a few fundamental assumptions which were structured as a rewrite system in computer terminology. Okay, so it basically is sort of being, you know, the algebraic thing. It works out to that, but it didn't start off with that. It just starts off with the concept of zero, and it starts off with the concept of some form of all that. It's a very simple one and it generates it very easily. We don't start with numbers or anything, we start with different rules. So I think that would perhaps be of interest. But I can't talk about it now because that would take too long. I'm going to pass over that because that will cause another argument. Right, so if we have, if we go, the part where we actually get this new potency, order 64, Dirac algebra. And one can split up the Dirac algebra into that format. We've got four complex scalars, twelve complex vectors, twelve complex fraternities, and thirty-six complex vector fraternities. So that's the Dirac algebra that can be split up as I use it. Now, when you count negative, it's twice. Yeah, yeah. Plus and minus counts as twice. Otherwise, it's 32. But I'm going to have to leave that. It's too much of an argument. So I'm saying that, I'm arguing here, that this is just a mathematical system, but to apply it to physics, you're trying to apply all the categories at once.

1:17:30 And you can kind of get away with that because you make it zero to an appropriate stage. And therefore, you're automatically including all the rest of the categories. But I don't want you to go into detail about it. just go past all this and this is how I now generate this so I've now started generating this from the rewrite system so I end up with this now the point is that this is a symmetrical to get onto the biology this is a symmetrical structure quite a symmetrical structure but one could generate the same thing more efficiently if you like by just using five operators instead of eight but these five are now no longer symmetrical so we have eight symmetrical operators, but now we've got five that are asymmetrical. And when you introduce fives, you start getting asymmetry. You lose the symmetry that you had with eights. Skip that. Skip that again. And I argued this in the previous talk. The combined Dirac state is a charge state as well as an energy state. and effectively we have that mapping for the charges and this charge has pseudo-scalar properties as vector properties this is what we were showing with the variance structure so we've got two lots of symmetry breaking at once so we're creating energy momentum mass in effect if you like and the other one is creating three different types of charge that we're making that five, we're breaking our symmetries. What do you mean by creation of concepts? Well, we're not really creating. I'm not talking about creating. I'm talking about creating from my starting point. I'm getting to a different concept. That's what I mean by creation. Generating, if you like, developing. It's not creating. It's not hard enough. Now, I think this is very important. The packaging process must occur as soon as we define the existence of the conserved quantity charge and the non-conserved entities as three-dimensional entities. And I said, in effect, it's only possible we conceive of a thing called conserved three-dimensionality. So there's a kind of thing which is a conserved three-dimensionality. It's a bit... So if you take angular momentum, that's a kind of conserved three-dimensionality. It's got another three-dimensionality other than the fact that it's a three-dimensional concept. It's got three attributes which determine its behaviour.

1:20:00 For example, it's got scale, size. it's got directionality and it's also got handedness so it's got three aspects and all conserved three dimensional quantities seem to have that kind of structure, a kind of broken symmetry, all conserved quantities like that, which one can represent by that kind of three dimensionality have a kind of broken symmetry in that three dimensionality, so as though we choose the first dimension for scale, the second one for dimensionality, and then the third one for the bookkeeper which has got nothing left to do except decide whether it's plus or minus or whether it's left-handed or right-handed. And one can see that pattern. I gave a talk about that a few years ago, a couple of years ago. So I can't really go into that now. So this kind of conserved three-dimensional quantity, so you get angular momentum, obviously they have the ordinary three-dimensionality, but that's within another structure of three-dimensionality. two three-dimensional structures here. We've got the one that's the P, that's the angular momentum, if you like, but we've also got another three-dimensional caused by these quaternion operators. And this, you get kind of scale, if you like, a scale term, or something like that. You get a dimensional term, and you get a bookkeeping term, which is that one. And again, you can do special relativity the same way, with proper time as the third one, potents as well. It might be special with rosavistic terms as nilpotents. In my opinion, there's no true four-dimensionality, because if you look at the Dirac equation, the time operator and the space operator both have a different operator in front of them, a different gamma matrix. So it's not a true four-dimensionality. It's only structured within something bigger, and if one writes it as a nilpotent, it's structured within a bigger three-dimensionality. space-of-time is truly four-dimensional, it's really sort of extra three-dimensionality. It's like a kind of double three-dimensionality. But you can't have curvature unless you have more three-dimension. Well, I'm not worried about curvature. Well, that's the reason why you don't need the fourth dimension. Okay. Well, I don't need the fourth dimension. I need three-dimension. That's the reason why, because you haven't got any curvature there. Right. Well, I can put curvature in, actually. I did that in the paper last year.

1:22:30 Well, you can look it up. I don't do it that way. I do a curvature. I do a matrix, right? You do a matrix. You don't just do a three-dimensional operator. You do three-by-three matrix. But you don't have... You do a one-plus-one system instead of three-plus-one. You can do it with one-plus-one. Oh, that's different. Yeah, okay, you believe that. But you only have one-plus-one at the quantum level, which I was interested in. So, if you define this as a nil potent, you've got to have at least one term, that's pseudo-scalar, because otherwise you won't zero it. So we need two three-dimensional structures. We need the quaternions for charge or something. We need those three quaternions. And we need the space three-dimensionality. So there's two three-dimensionalities in it. And we find that pattern repeats all over the place. And so the pseudo-scalar term becomes the bookkeeper. it's the energy, the weak charge it's the one that does the aggregation and dispersion, it's the one that determines asymmetry asymmetry always and I say it's part of an incomplete quaternion set it's also mathematically ambiguous to pseudoscalar quantity so we can't distinguish between plus and minus sides and we so we need, as it were 2.5 Paternian series to actually do physics so we can actually do physical information with only 2.5 of these infinite sets so we can't do mathematics for that full one now it seems to me that we've got similar structures in biology you've got possibly for slightly different reasons we don't know yet, we'll have to look at it So, I'm going to just very briefly cover a few topics that will relate to. First of all, relevance of platonic solids. Five-fold broken symmetry in Fibonacci series, double helix-type idea, and 64 units. And one will see similar patterns. One can argue about the reasons for those, but maybe for some sort of mathematical structure which somehow is relevant to both. But as I say, this is exploratory, so I don't expect perfect.

1:25:00 explanations of anything so I think there are four fundamental parameters in the Ferminic state a slider showed earlier showed what I call a dihedral symmetry which is why I asked Clive the question much earlier whether one could have an anti anti-indistinguishable because I've got one distinguishable and two anti-indistinguishables for each of these parameters. And so we can actually represent them either as a cube or a tetrahedron. So we've got the properties and antiproblems that really matter, conserved, non-conserved, discrete and continuous. So let me show you how one can represent those fundamental properties in that way. And you see I've got my two... On each case I've got... One of them is indistinguishable and the other two are anti-indistinguishable in that terminology, which is why I asked you that question. That's the second time these things have come up. I've shown this a million times. It's always been part of my work. And what worries me about the right-hand colony is that whereas time is stated as continuous, space is discrete. Well, it is. It's three-dimensional. It is discrete. It's discrete, but it's not conserved, you see, so we don't have a fixed discrete lesson. Well, space is discrete whereas time continues. Zeno's paradox, no? Sorry? It seems you're using discrete to mean. No, I think I mean it in a different way to the way people do. I understand space really to be structured by sort of non-standard analysis and time to be standard analysis. So I understand space to be... I'd say if one can measure something and one can divide it up, then it's discrete. it's not a fixed discreteness because it's non-conservant but the only way of saying that space is discrete is to say that you can only operate you can only occupy certain points and you can't occupy no, no, no, not at all you can at any moment you always divide your space up always, but you change that division, you can change it infinitely so the fact that doesn't make it continuous, it makes it discrete. I think we've discussed, Lou and I have discussed

1:27:30 this. I think we have. You make something infinitely divisible, it's not... That wouldn't necessarily create any stability. If you make something infinitely divisible, it's discrete, it's not continuous. Lou's point was that you don't need continuous time in that sense. I do need continuous time. Because I don't... If time can't run backwards, then it's continuous. that's my argument if time is continuous it can't run backwards nor can it branch off all discrete quantities in my thinking it's always got this anti-community property well that's fair enough I have a stronger role but I do have a reason it's a case where the rest of the world continuous. I don't know why. Because the definitions they give. Okay, but I use continuous to mean something different. I use continuous to be absolute. Okay, thank you. Okay, I'll discuss that with you. I can do a cubicle representation. I can actually represent my properties. I can represent each parameter, say, as one of these red lines. It's easier, actually, if we see it like that. I can represent each parameter as one of these lines like that, going to one of the corners, and each of these three dimensions has one of those properties or anti-properties, so plus and minus x, plus and minus y, plus and minus z sort of thing, and the cyan lines represent basically a reciprocal group, not a reciprocal group, a dual group, which to me is what I understand to be one can use that as a representation of vacuum states, for example one can also use a tetrahedron to represent it or one can use a scar tetrahedron so this is one can represent the physics as I do it by a tetrahedron same thing, one can represent the the vertices or the faces what am I doing there the four vertices, how many faces are there four faces, four faces four vertices, four faces parameters with three properties coming into it and you see one of the properties will be the same in each indistinguishable and two of them will

1:30:00 be anti. This is very old in my work. Yeah, I'm interested in that. Yeah, I know but people keep interrupting me so how can I continue? So, it needs eight primitive units and these are only five composite ones. So, we're now breaking the symmetry. This is the second point, broken symmetry. Five composite ones, eight primitive ones. So, the most efficient structure is not the most primitive. It's an argument very similar to Higgs mechanism. And if we flatten the tetrahedra, we can use it as kites and darts and then we can structure as a Penrose tiling if we choose. And if we're representing no potents this way we can use a... we say they're not repeatable the pattern. But you haven't got on to biology. No, I'm not going to do the biology. So Fibonacci series is used in the Penrose tiling. I can get into it very quickly. So as we'll point out platonic solids include pentagonal symmetries so if you use cubes and so on in it. And the symmetry breaking term is always the fifth one, if you like, to time the energy of the weak charge of symmetry. And I was recently at a conference where this man Johansson showed a method of excusing Fibonacci as the method of information processing. He's actually been, used the universal rewrite system involved in that. And I noticed his algorithm was eights split into fives and threes. And that's how he did his algorithm. That's just quite interesting. Let me just point out, Clifford Algiers, of course, allows to use multiple dimensionalities. And I think there's only one type of dimensionality, ultimately, three-dimensionality. I found a biological example of using double three-dimensional space. The Eilert showed that the minimum reputation of the seashell was doubling with three-dimensional space. and it seems that this development needs it's because you need a development in time to do that so this makes sense in terms of what I said is that to structure time into it you need a three dimensional operator not a four dimensional one

1:32:30 you need a four dimensional operator to structure time in relation to space we might be able to say tetrahedral arrangement is the most efficient And if we let that structure retain our density, then that will require three-dimensionality extra to structure it. You can't structure three-dimensional space with vector units. No, but you can structure a spiral with them, can't you? Which is what we're going to do. And now, an argument I've long had is that nil-potency squares it to zero, and effectively a fermion doesn't, what it's really doing is saying that the rest of the universe vacuum zeros it it only exists in relation to the rest of the universe which is what I can understand vacuum to mean so it only remains in terms of its other half and I think that's the reason why we get half spin the peculiar half spin, it's a bit like Newton's third law I'll have to skip that what is the reason we get half spin? well because the fermion doesn't exist so it has to structure itself it can't exist if it doesn't create vacuum at the same time and therefore it exists at the same time with its vacuum I could almost agree with that more than agree with that ok well that's its surprise that makes it sort of double helical if you like I might try to use the word double helic but the fermion and vacuum itself are together which is why it has this peculiar arrangement for its spin The way I do weak contraction, it becomes a harmonic oscillator, so it's always a dipole. Now, a dipole term in physics tends to introduce a 1 over r to the 4th dependency, exactly like vacuum term, exactly like van der Waals. And in planetary orbits, for example, 1 over r to the 4th term produces a spin term. So I can see a pattern there. And, of course, the van der Waals force, in principle, is the aggregation of matter term. I think we'll just leave out the left hand bias can't do anything about that and then of course there's 64 units in the amino acid triplets 64 units of Dirac algebra is there a connection? one can