Bombadil Posted July 17, 2008 Report Share Posted July 17, 2008 They are fine but they don't answer my original question. I have no idea as to why you should want this “hyperplane” as part of our discussion. I suspect that it is a misunderstanding as to what a boundary is I was under the impression that the two ideas where related. I am not sure here but I definitely get the impression that you seem to miss the point of algebraic substitution. All that I am saying is that we should decide on which way to write the equation and not constantly change the way we write it. It just seems more convenient to stick with one way of writing it so as to avoid confusion as to why someone is writing it differently. Your concept of rotation is totally dominated by your intuitive understanding of three dimensional relationships. As I said to you long ago, the human mind has no experience with four dimensional relationships so we must be very careful. We cannot use our intuition at all and must depend upon analytical geometry, not mental pictures. How is the point at which the rotation angle is measured defined and is such a point always stationary or can such a point move during a rotation? Also are there any other properties that would seem that they would hold for a rotation that don’t hold in more then 3 dimensions? So in your proof one of the lines that define the plane of rotation is orthogonal to all previous planes of rotation (this line is in fact the axis just added) while the second line defining the plane is the line on which the projection is made onto. This results in a scaling of all of the points except one when a rotation is performed. Quote Link to comment Share on other sites More sharing options...

Doctordick Posted July 27, 2008 Author Report Share Posted July 27, 2008 I suspect that it is a misunderstanding as to what a boundary is I was under the impression that the two ideas where related.Boundaries are not an important issue in my presentation; I was only pointing out how the issues of significance change with increasing dimensionality. In a case of [imath]n=10^{100}[/imath] there exists no way of naming or even describing the collection of significant k-simplexes which exist in an n-simplex of interest.All that I am saying is that we should decide on which way to write the equation and not constantly change the way we write it. It just seems more convenient to stick with one way of writing it so as to avoid confusion as to why someone is writing it differently.And my point would be that there is no confusion if what the equation is saying is understood.How is the point at which the rotation angle is measured definedRotation angle is not defined “at a point”. Rotation is defined by a plane (or two non-parallel lines). The easiest way of defining the rotation angle is to specify a specific line in the structure being rotated and compare that specific line with its position after the rotation. The line before the rotation and the same line after the rotation define the plane of rotation and the angle between those two lines is the angle of rotation (in that plane we are dealing with ordinary plane geometry).So in your proof one of the lines that define the plane of rotation is orthogonal to all previous planes of rotation (this line is in fact the axis just added) while the second line defining the plane is the line on which the projection is made onto. This results in a scaling of all of the points except one when a rotation is performed.Yes! Quote Link to comment Share on other sites More sharing options...

Bombadil Posted August 1, 2008 Report Share Posted August 1, 2008 There is an interesting corollary to the above proof. Notice that the rotation specified in the final paragraph changes only the components of the collection of vertices along the x axis and the nth axis. All other components of that collection of vertices remain exactly as they were. Since the order used to establish the coordinates of our polyhedron is immaterial to the resultant construct, the nth axis can be a line through the center of the polyhedron and any point except the first and second (which essentially establish the x axis under our current perspective). It follows that for any such n dimensional polyhedron for n greater than three (any x projection universe containing more than four points) there always exists n-2 axes orthogonal to both the x and y axes. These n-2 axes may be established in any orientation of interest so long as they are orthogonal to each other and the x,y plane. Thus, by construction, for any point (excepting the first and the second which establish the x axis) there exists an orientation of these n-2 axes such that one will be parallel to the line between that point and the center of the polyhedron. Any rotation in the plane of that axis and the y axis will do nothing but scale the y components of all the points and move that point through the collection, making no change whatsoever in the projection on the x axis. So in order to generalize this to a projection on more then one dimension, say two dimensions, all that we have to do is define one of the lines to be perpendicular to all the previous planes of rotation plus perpendicular to the plane we are projecting onto (this is still the last axis added), while the second line is an axis of the plane we are projecting onto. This changes the projection along one of the axis’. We then repeat this using the second axis of the plane that we are projecting onto for the second line, this adjusts the second axis of the plane we are projecting onto. While this seems that it should move the point to where we want to move it while only scaling the remainder of the points, it seems to me that both measurements on the plane being projected onto must be defined in the points making up the projection. Even with doing this, it seems that moving the projection of n-1 of the points will completely define the projection of all n of the points on the n-simplex so that we can’t move the nth point to any location that we want and that it will have only one or perhaps two locations that it can be projected onto. Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 3, 2008 Author Report Share Posted August 3, 2008 So in order to generalize this to a projection on more then one dimension, say two dimensions, all that we have to do is define one of the lines to be perpendicular to all the previous planes of rotation plus perpendicular to the plane we are projecting onto (this is still the last axis added), while the second line is an axis of the plane we are projecting onto.Your picture of what is going on might be somewhat confused though you certainly seem to have comprehended the issues. Let me try to go through the steps from a slightly different perspective. At least the first step of adding a “y” axis and creating a two dimensional projection. We begin by finding an orientation which will make the points on our n-simplex project on that “x” axis to the distribution we desire. (That we could do this was proved in the original theorem.) Now, choose another axis orthogonal to “x” to be our “y” axis. Our n-simplex is defined by n points which lie in an n-1 dimensional space space. It follows that there are n-2 axes orthogonal to that “x” axis which include our new “y” axis. Thus there are n-3 axes orthogonal to both axes. Since we are discussing rotations which do not effect our x distribution and provide an ability to adjust the distribution on the “y” axes, the planes of rotation we are interested in are defined by that “y” axis and one of those n-3 axes orthogonal to it. We can begin by enumerating the n points to be distributed. We then pick the first point and observe the projection on the “y” axis. No rotations are required because one point does not a pattern make. So we pick the second point. Once again, no rotations are required as two points can provide nothing more than a scale and, having a scale, there is nothing to measure. When we pick our third point, we are required to find a rotation which will scale the points already established and move the third point through that set. The plane of rotation is defined by the “y” axis and the line orthogonal to that axis going through that next point. We clearly have n-3 possible rotations one of which will be the one specified by that third point. Each time we pick a point and apply the appropriate rotation, the number of points to be moved and the number of rotations available to cause that move both decline by one. When i is 3 (we have n-2 points to distribute) we only have n-3 rotations available. So, as you have mentioned, after we use up all of those rotation, we are left with the projection of one point not yet properly positioned (that last nth point). We need one more axis orthogonal to the “y” axis in order to provide us with this last rotation. We can pick up this axis by using the “x” axis. We only need to assure that the rotation does not change the actual distribution already established on both of these axes, though that rotation can be allowed to scale that distribution. All we need to do is rotate in the x,y plane (an act which makes no change in the distribution so far obtained as our distribution is now an established two dimensional distribution) until the "y" axis is parallel to the line between the current position of the nth point (where it is currently projected in the x,y plane) and the point to which we wish it to be projected. At this point, the line to the final point is orthogona to all the lines to the other points projecting onto the desired x,y plane and any rotation orthogonal to this new "y" axis will not change the "x" position of the projection of that nth point. We now have that final plane which can be used to position the nth point. Rotation in that plane will merely scale the projection in the direction that point is to be moved while it moves that final point through the rest of the distribution. The addition of a "z" axis ends up leaving us short two rotations; however, we then have two planes (the z,x and the z,y) which can be used to reorientate the "z" axis and provide those missing rotations. Think about that for a while. Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Bombadil Posted August 9, 2008 Report Share Posted August 9, 2008 We can begin by enumerating the n points to be distributed. We then pick the first point and observe the projection on the “y” axis. No rotations are required because one point does not a pattern make. So we pick the second point. Once again, no rotations are required as two points can provide nothing more than a scale and, having a scale, there is nothing to measure. When we pick our third point, we are required to find a rotation which will scale the points already established and move the third point through that set. The plane of rotation is defined by the “y” axis and the line orthogonal to that axis going through that next point. We clearly have n-3 possible rotations one of which will be the one specified by that third point. Then the n-3 rotations are defined by the y axis and a line orthogonal to the x and y axis? Such a rotation will only scale the y axis and move one point and will not alter the x projection of the n-simplex. Don’t we have to define the scale with two points that nether of which lie on the n-3 axis’s? If so, isn’t it possible that all such points will lie at the same y coordinate? In which case, how can we use it for a scale as it would put all points at a infinite distance from them. So, do we have to choose the y axis so that this doesn’t happen? All we need to do is rotate in the x,y plane (an act which makes no change in the distribution so far obtained as our distribution is now an established two dimensional distribution) until the "y" axis is parallel to the line between the current position of the nth point (where it is currently projected in the x,y plane) and the point to which we wish it to be projected. This rotation in fact scales both the x and y axis? If so is it moving the projection of the second point that was added (which would be the point that was used to define the scale on the x axis)? If I understand this right, the points that are used to define the scale on both axis’s have to be separate points? If so is the point that we have to use to define the scale on the y axis uniquely defined by the y axis? Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 9, 2008 Author Report Share Posted August 9, 2008 Hi Bombadi, nice to see you are thinking these things out.Then the n-3 rotations are defined by the y axis and a line orthogonal to the x and y axis? Such a rotation will only scale the y axis and move one point and will not alter the x projection of the n-simplex.That is correct.Don’t we have to define the scale with two points that nether of which lie on the n-3 axis’s?I think you are overlooking the procedure being used. First we decide the order with which we are going to deal with the n points (we ignore the current projection of all the points on the “y” axis except the one of interest). We then accept point #1 wherever it happens to project. (You do realize that the original distribution on the “x” axis has been established; i.e., we have rotated the n-simplex to some specific orientation and, that orientation yields some projection on the “y” axis.) Essentially, we do not wish to effect the distribution on the “x” axis so there are only n-3 possible rotational planes (the rotational plane is defined by two axes, one of which must be the new “y” axis, which is orthogonal to the “x” axis and the remaining n-3 can be any orthogonal set orthogonal to “x”). So point #1 projects wherever it projects (no rotations are required). Then we pick point #2 which also projects wherever it projects (and no rotations are required because the best it can do is establish the scale and there is nothing to measure). Now, you bring up the point that it is indeed possible that the second point projects to exactly the same point. That case is indeed possible but it means that the line between point #1 and point #2 in the original n-simplex is exactly orthogonal to our new “y” axis. If that is the case, point #2 was a poor choice; choose another point as point #2. Well, you say, being the observant thinker you are, suppose all the remaining n-1 points satisfy that same constraint (the line between point #1 and point #2 is orthogonal to the selected “y” axis). It turns out that specific case cannot occur because there are a lot more point pairs than there are orthogonal dimensions. As a matter of fact, the worst case scenario is actually the orientation I used in the original proof of my theorem (the nth point was placed on an axis orthogonal to all the dimensions used to describe the (n-1)-simplex). Thus it is that we know there exists at least one pair of points which project to different positions on the “y” axis (there is no orientation of our n-simplex which projects every vertex to the same point). So it is just a point of proper selection of the first two points (just don't select two which project to the same point on the “y” axis).So, do we have to choose the y axis so that this doesn’t happen?No, you need to choose the first two points so this doesn't happen. After that, we can use rotation to adjust the distribution so in all other cases, we are not bothered by the new point projecting to the same point as one already distributed as we are going to move it anyway.This rotation in fact scales both the x and y axis? If so is it moving the projection of the second point that was added (which would be the point that was used to define the scale on the x axis)?You are missing a very important point. The issue of scale is that the scale has to be determined from the distribution of points by some algorithm on the distribution itself. The only reason I brought up scale at the point where the distribution had only two points is that, if you have only two points, the only algorithm for determining scales is “the two points”. As I said, the rotation in the x,y plane does not change the distribution so far obtained in any way; except for that last point, every point has been moved to the position desired. This distribution is some scaled version of the distribution you desired. The rotation merely changes the representation of that distribution. After establishing the new “y” axis (and the new “x” axis) the scale in those directions has to be established by some procedure based upon the distribution in those directions (they can be seen as independent coordinates). Having established that scaling, however it is done, the final rotation only scales the distribution in the new “y” direction while it moves that last point through the distribution. If I understand this right, the points that are used to define the scale on both axis’s have to be separate points? If so is the point that we have to use to define the scale on the y axis uniquely defined by the y axis?The procedure used to define the scale in a distribution of, say [imath]10^{100}[/imath] points could be quite complex but it can only be established by some procedure involving those points only (there is no outside source we can use to establish a scale). For example, thinking of our projection of this n-simplex as a TOE for our known universe, there are a large number of reference points (referred to as scratches) on a complex structure called a platinum bar found within another complex structure called Paris which is used to establish scale in this TOE. Or perhaps the algorithm has been changed to another rather complex collection of events (called the wave length of a photon from a specific transition). Scale is scale and it has to be established somehow. It is an assumption that this scale establishing procedure is independent of actual axes directions used but, when ever we go to actually establish a measurement, we are doing it only in one dimension. Life is full of subtlety people seldom think about. Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Bombadil Posted August 15, 2008 Report Share Posted August 15, 2008 I didn’t get to posting this earlier but there are still a few things that I’m wondering about although I think that I’m getting the basic idea. Well, you say, being the observant thinker you are, suppose all the remaining n-1 points satisfy that same constraint (the line between point #1 and point #2 is orthogonal to the selected “y” axis). It turns out that specific case cannot occur because there are a lot more point pairs than there are orthogonal dimensions. As a matter of fact, the worst case scenario is actually the orientation I used in the original proof of my theorem (the nth point was placed on an axis orthogonal to all the dimensions used to describe the (n-1)-simplex). Thus it is that we know there exists at least one pair of points which project to different positions on the “y” axis (there is no orientation of our n-simplex which projects every vertex to the same point). So it is just a point of proper selection of the first two points (just don't select two which project to the same point on the “y” axis). So we have the projection of the first point (call it point #1) and there are n more points that we can choose as a second point. If we choose one (call it point #2), in order for them to project to the same point on the y axis we know that a line passing through point #1 and #2 has to be perpendicular to the y axis. We also know that any line that passes through point #1 and one of the n-1 remaining points must be perpendicular to the y axis in order for it to project to the same y coordinate as point #1 and #2. But there are only n-2 more perpendicular dimensions not including the y-axis while there are n-1 more points to project so at least one of the points must project to a different y coordinate. This seems to lead me to the question, how few points can we project onto? It seems obvious that we can project down to two points, that is we can not only project the n-simplex into any combination of n points we also can project onto any number of points m where m is the number of points and [imath]2\leq m\leq n[/imath]. Now, if we want the projection to move from one projection to another in a particular way can we simply perform multiple rotations at the same time to obtain any change in the projection? Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 16, 2008 Author Report Share Posted August 16, 2008 This seems to lead me to the question, how few points can we project onto? It seems obvious that we can project down to two points, that is we can not only project the n-simplex into any combination of n points we also can project onto any number of points m where m is the number of points and [imath]; 2leq mleq n[/imath].You are correct, it is impossible to project every vertex of an n-simplex to a single point on a line; however, I contend that this is not a violation of my proof under the definition of “every conceivable distribution” as “one point” is not technically “a distribution of points”. The assertion which will be proved is that every conceivable distribution of points on the line segment is achievable by a specifying a particular rotational orientation of the polyhedron relative to the line segment of interest.On the other hand, considering the scale issue, the problem of projecting every vertex of an n-simplex to a single point is nothing more than setting the scale factor desired to zero. In any case, it is a rather trivial issue of little serious use. Now, if we want the projection to move from one projection to another in a particular way can we simply perform multiple rotations at the same time to obtain any change in the projection?First, I wouldn't use the term “multiple rotations” as any n dimensional rotation can be seen as a collection of rotations in those different planes referred to: i.e., I am using the word rotation to refer to the complete set of defined rotations leading to a particular orientation (there are a great number of paths to that orientation). In any case, here it is best to think in terms of differential changes in that projection. A specific differential change in that projection (and that would be a simultaneous specific differential change in position of every point in that distribution) would essentially define a specific differential rotation (a specific reorientation). Now rotation has a very strong connection to quantization. The fact that rotation through 360 degrees brings us exactly back to the original orientation yields quantized angular momentum. That is to say, if the n-simplex is taken to be an entity who's motion is described by quantum mechanics, one would expect each and every one of those rotations to conserve angular momentum (it's a solid object and there is nothing outside itself to influence its motion). Another way to view that issue is to realize that, if the current projection and the differential change in position are both known, all future projections are known. There are two specific difficulties in that picture. First, if it is an entity who's motion is described by quantum mechanics, there is an uncertainty relationship between orientation and angular momentum which, at best, says that all you can do is come up with probability distributions. And second, as a TOE, though it has been proved to be perfect, it is quite useless for making predictions as it tells us essentially nothing about what to expect (as who among us can describe that rotation, much less the projections which are to be expected). That is exactly where my fundamental equation comes into play. It turns out that one can show that this rotating n-simplex is exactly equivalent to the solution to my fundamental equation when the dimensionality of the equation is taken to n, the number of arguments of [imath]\vec{\Psi}[/imath]. (It's not a trivial issue and it won't appear in my presentation until quite a bit later but I will get to it eventually if I can get general comprehension out there.) Meanwhile, you might keep in mind that the professionals have already decided I am a crackpot so don't get too excited. Do you have any understanding of my deduction of that fundamental equation? I notice you haven't seemed to take much interest. Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Bombadil Posted August 21, 2008 Report Share Posted August 21, 2008 First, I wouldn't use the term “multiple rotations” as any n dimensional rotation can be seen as a collection of rotations in those different planes referred to: i.e., I am using the word rotation to refer to the complete set of defined rotations leading to a particular orientation (there are a great number of paths to that orientation). In any case, here it is best to think in terms of differential changes in that projection. A specific differential change in that projection (and that would be a simultaneous specific differential change in position of every point in that distribution) would essentially define a specific differential rotation (a specific reorientation). Now rotation has a very strong connection to quantization. The fact that rotation through 360 degrees brings us exactly back to the original orientation yields quantized angular momentum. That is to say, if the n-simplex is taken to be an entity who's motion is described by quantum mechanics, one would expect each and every one of those rotations to conserve angular momentum (it's a solid object and there is nothing outside itself to influence its motion). Another way to view that issue is to realize that, if the current projection and the differential change in position are both known, all future projections are known. From what I can tell the meaning of it being quantized means that only energy levers that satisfy a certain constraint are allowed. So in the case brought up in “Some subtle aspects of relativity” it would mean that only certain momentum levels are allowed? In this case I’m not so sure if I understand you right, it has something to do with only allowing a full rotation but I don’t know in what way. Isn’t for all practical proposes though, a rotation through 360 degrees impassable in that it requires one of the elements to go to infinity? That is exactly where my fundamental equation comes into play. It turns out that one can show that this rotating n-simplex is exactly equivalent to the solution to my fundamental equation when the dimensionality of the equation is taken to n, the number of arguments of vec{Psi}. (It's not a trivial issue and it won't appear in my presentation until quite a bit later but I will get to it eventually if I can get general comprehension out there.) Meanwhile, you might keep in mind that the professionals have already decided I am a crackpot so don't get too excited. This whole thing kind of reminds me of some of the theories that I have heard of before, although without a better understanding of them I have no idea if there is any link between them and what you have derived or if they’re just working on a completely different construct. As for them labeling you a crackpot I’m really not concerned about it. Do you have any understanding of my deduction of that fundamental equation? I notice you haven't seemed to take much interest. I’m really not sure what it is that you are referring to unless you are referring to the fundamental equation [imath] \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi} = K\frac{\partial}{\partial t}\vec{\Psi} [/imath] in which case I was under the impression that we had covered the important issues that need to be understood. The other thing I see as a possibility is you are referring to the equation you bring up in post 21 of this thread, but I was under the impression that we needed to cover relativistic effects before we could derive that equation. Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 22, 2008 Author Report Share Posted August 22, 2008 From what I can tell the meaning of it being quantized means that only energy levers that satisfy a certain constraint are allowed. In a sense, yes. The problem is that there is a lot of common quantum mechanics which you do not know. If the wave function (in this case [imath]\vec{\Psi}[/imath] must return to exactly it's value over some specified range then its wave length must divide evenly into that range and that constrains the momentum to specific values. It is very analogous to the strings on a stringed instrument, the strings vibration must be exactly zero where it is tied down and this limits the possible vibrations to a limited set (often referred to as the fundamental and its harmonics). The energy is related to the frequency and, in those instrument strings, the wave length is fixed but the frequency is set by the tension and the weight of the strings. So in the case brought up in “Some subtle aspects of relativity” it would mean that only certain momentum levels are allowed? In this case I’m not so sure if I understand you right, it has something to do with only allowing a full rotation but I don’t know in what way.You are mixing two different threads here and two different quantization issues. In this thread, I am talking about the quantization required by cyclic repetition in position (a full circle) where, in the other thread, I am talking about the consequences of having a fixed wave length. In that case, there is nothing to force that fixed wave length but luck; however, once it has been established, there is no observable mechanism to change it and that is another long discussion.Isn’t for all practical proposes though, a rotation through 360 degrees impassable in that it requires one of the elements to go to infinity?Here you are underestimating the subtleties of the possible rotations of an n dimensional object. Take a careful look at that diagram I made for Turtle (post #15). Take the two axes I define there and squeeze them towards one another a bit then picture what happens if the 4-simplex is rotated in opposite directions on both axes at the same time. All that happens in the projection is that the points kind of wobble in their positions.This whole thing kind of reminds me of some of the theories that I have heard of before, although without a better understanding of them I have no idea if there is any link between them and what you have derived or if they’re just working on a completely different construct.I doubt anyone has considered exactly what I have considered.I’m really not sure what it is that you are referring to unless you are referring to the fundamental equation [imath] \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi} = K\frac{\partial}{\partial t}\vec{\Psi} [/imath] in which case I was under the impression that we had covered the important issues that need to be understood.I was referring to that equation but I certainly have not covered the derivation in this thread, Read the note in my response to you in the “Some subtle aspects of relativity”. If you feel that you understand the derivation of that equation, then don't worry about it; however, I suspect there are aspects of that derivation which are not clear to you. If I am wrong than you are the only person I know who has managed to understand what that equation is based upon. (Except Anssi of course, I think he has a good idea as to what it is based upon.) Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Rade Posted August 25, 2008 Report Share Posted August 25, 2008 DD, I have a question. Would it be correct to say that an ontological element used as the existential basis of your fundamental equation must have the geometry of an n dimensional tetrahedron, where n >= 4, since it is only this value of n that leads to an undefined object ? Quote Link to comment Share on other sites More sharing options...

AnssiH Posted July 5, 2011 Report Share Posted July 5, 2011 Started reading this and just spotted two things so far; The new radius will be given by the square root of the sum of the old radius squared and the distance the old polyhedron was moved up in the new dimension squared. That is exactly the same distance the new point must be from the new center. I found that part a bit sloppy because I first read it as saying "...the distance the old polyhedron was moved up in the new dimension squared is exactly the same distance the new point must be from the new center" which of course wouldn't make any sense. I realized what you meant was "the new radius... ...is exactly the same distance the new point must be from the new center" (which is kind of tautologous :) Overall a lot of that explanation would be easier with couple of pictures, it's a bit cumbersome sometimes to understand it in text. [math]r_{up} = \frac{1}{\sqrt{2n(n+1)}}[/math] You have a small typo there, I believe that should be [imath]x_{up}[/imath]. -Anssi Quote Link to comment Share on other sites More sharing options...

Doctordick Posted July 5, 2011 Author Report Share Posted July 5, 2011 Started reading this and just spotted two things so far;I had a hell of a time finding "...the distance the old polyhedron was moved up in the new dimension squared is exactly the same distance the new point must be from the new center" which of course wouldn't make any sense.until I realized you omitted, "The new radius will be given by the square root of the sum of the old radius squared and ...", which is a pretty important part. I have made a minor change in the wording which may or may not make the issue clearer. The other point is absolutely correct; however, when I edited the thing the new system stripped out all the back slashes so I had to fix everything. I wish they would fix that kind of thing. I also made some pictures of the constructs required. I only made the first two constructs as, all that is left is the third (which would require me to make a projection of a three dimensional object which I don't feel qualified to do a good job) and nothing more as everything else is essentially outside our ability to picture (requires picturing 4 dimensions). The actual construct is no more than following exactly the prescription constructing the n=2 case. That is far more simple than trying to picture it. I will send you via e-mail the pictures of the construct as I made them. Let me know what you want me to do and I will do it. You might post those pictures with your comments if you want. Sorry I am so slow at getting back to you but I am an old man and I don't work quickly any more. I used to think that the great advantage of getting old is that you know so much more than you did when you were young -- but then you start to forget. I had a hell of a time understanding what I was doing in that post. Thanks for your efforts -- Dick Quote Link to comment Share on other sites More sharing options...

AnssiH Posted July 6, 2011 Report Share Posted July 6, 2011 I have made a minor change in the wording which may or may not make the issue clearer. It's clearer now. It might be beneficial to future readers if the images you sent could be embedded to appropriate places in the OP, it just makes it more trivial to read the thing. It would need to be uploaded elsewhere for that, maybe to http://tinypic.com/. Or maybe just add the files as attachments to the OP. In the meantime; [math]r_n = \sqrt{x_{up}^2 + r_{n-1}^2}[/math] and [math] 1 = \sqrt{r_{n-1}^2 + (x_{up} + r_n)^2 }[/math] That all looks valid... The solution of this pair of equations is given by [math]r_n = \sqrt{\frac{n}{2(n+1)}} [/math] and [math]x_{up} = \frac{1}{\sqrt{2n(n+1)}}[/math] ...and so does that after I've just been playing around with your solutions, but still I could not actually figure out what are the algebraic steps that got you to those solutions. It's just my lack of experience with math again, I'm sure the path is fairly obvious once you point it out to me... :I So if you could just explain it a bit, while I make the necessary preparations to kick myself. -Anssi Quote Link to comment Share on other sites More sharing options...

AnssiH Posted July 7, 2011 Report Share Posted July 7, 2011 Also, in the meantime, I read through the rest of the OP, not with great detail but enough to understand what you mean. I don't have any problems with it, it all seems quite valid. Essentially each new point entails a new orthogonal coordinate axis for the polyhedron, which also makes it possible to control the position of its projection independently from the other projected vertices, via appropriate rotation on the polyhedron. And furthermore, because the information is embedded to the actual distribution of the points, it is always possible to use up to 3 projection axes without giving up any possibilities in terms of representing any sort of distribution. I'm still thinking about the details on that one but so far it seems valid to me. I'm not sure what to make of this result. Maybe it implies that higher (than 3) dimensional representations of information are less general (in terms of ease of representing any information), or maybe there's a point to be made about how the higher dimensional projections wouldn't (perhaps) have independent controls for all the elements. Or maybe it means nothing. I can't really pick up much right now. Do you have any additional thoughts about it yourself? -Anssi Quote Link to comment Share on other sites More sharing options...

Doctordick Posted July 10, 2011 Author Report Share Posted July 10, 2011 Thanks Anssi, I appreciate your reading that thing through. I don't expect you to follow this post but rather put it forth for interests sake. I'm not sure what to make of this result.Well, if you were to take the standard position held by the scientific community (i.e., that an explanation which explains everything known without a flaw of any kind is a correct explanation of reality) then it must be that the universe IS a rotating minimal n-dimensional polygon with unit edges. That is about the simplest explanation I have ever heard and, if you can't prove it is wrong, by Occam's principle, it certainly stands above any other “theory”. To quote Wikipedia, “in the scientific method, Occam's razor is not considered an irrefutable principle of logic, and certainly not a scientific result”, however, if one can prove it is error free, it seems to me that the issue of irrefutable logic can sort of be laid aside. I will tell you how I was first led to the necessity of that result. If you look at my deduction of Schrödinger's equation you will notice that I divide the labels “x” into three independent sets which I call x, y and z thus leading to a three dimensional picture (the hypothetical tau is still necessary for exactly the reason it was originally necessary). This was done because Newton's equations are rather useless in a one dimensional picture; a little better in two dimensions but really valuable in three dimensions. (Perhaps the reason we think we exist in a three dimensional universe: i.e., it is the simplest picture which has any real value to our survival. What immediately occurs to any thinking person at that point is, why not look at a four dimensional representation of reality. The problem with a four dimensional representation is that it is, as far as I am aware, impossible to picture in our minds eye (remember, it's five dimensional after we add in that tau). Thus problem must be examined from an analytical perspective and it is not easy to express how it would appear to the proper sensing organ. If we have to do the problem analytically, why stop at four. It is entirely possible that some higher dimensional representation could yield a more rational picture. In fact, why not view the universe as one point in an n dimensional Euclidean space: that is, if we have n pieces of data, why not an n dimensional model. One immediate advantage of using an n dimensional model to represent n numbers in the observation is that we no longer need to deal with the problem of two knowable events falling on the same point; note however that we still need a background of unknowable data to constrain the knowable event to actual observation. Clearly, since we have only one knowable point and our fundamental equation is valid only in the center of mass system, the probability function which represents our event must be symmetric about the origin (or we can move the origin to make the statement true). The unknowable data can be viewable as forming a potential well which constrains our point to the origin: by simple symmetry it cannot depend on any angles. The interesting thing about this model is that the radial function, the only part where any variation is possible, is totally unimportant. Clearly, any solution for the entire universe must be dimensionally scalable and we only have one linear dimension: i.e., "r". In this view, the fundamental equation of the universe (seen as a point in an n+1 dimensional sphere, n angles and one radius) becomes [math]\left\{\vec{\alpha}\cdot\vec{\nabla}+\beta g®\right\}\vec{\Psi}=k\frac{\partial}{\partial t}\vec{\Psi}=ikm\vec{\Psi}[/math] where we have explicitly inserted conservation of energy. Consistent with previous practice, we may use the implied operator identity [math]\vec{\alpha}\cdot\vec{\nabla}+\beta g®=ikm[/math] to generate the n-dimensional Laplacian equation [math]\left\{\nabla^2+2g^2®\right\}\vec{\Psi}=-2k^2m^2\vec{\Psi}=-K\vec{\Psi}[/math] Note the two arises because of the fact that [math]\alpha^2=\frac{1}{2}[/math]. The curious thing about the equation above is that the angular part, the only significant part, admits of solution in closed form. Note that the (n+1) dimensional Dalembertian operator (that would be n angles and one radial measure) may be written [math]\nabla^2=\sum_{i=0}^n\frac{\partial^2}{\partial x^2}= \left(\frac{1}{r}\right)^n\frac{\partial}{\partial r}r^n\frac{\partial}{\partial r}+\frac{1}{r^2}\sum_{i=1}^n\left(\prod_{l=1}^{i-1}csc^2\theta_l \right)(csc\theta_i)^{n-i}\frac{\partial}{\partial \theta_i}(sin\theta_i)^{n-i}\frac{\partial}{\partial \theta_i} [/math] essentially written in hyper spherical coordinates where [math]r=\sqrt{\sum_{i=0}^n x_i^2}[/math] [math]\theta_{i+1}= arccos\left(\frac{x_i}{\sqrt{\sum_{j=i}^n x_j^2}}\right)[/math] and [math]x_i=rcos\theta_{i+1}\prod_{l=1}^i sin \theta_l[/math]. Note that theta sub n is defined to be zero otherwise one ends up defining an [math]x_{n+1}[/math], a slight inconsistency since our n+1 arguments include “r”. Substitution (I don't expect you to confirm this as the math is not trivial) will confirm that the fundamental equation in this form has a general solution of the form [math]\Psi=\frac{1}{\sqrt{r^n}}U®\prod_{i=1}^n\Phi_i(\theta_i)[/math] which is essentially the form of an n dimensional angular momentum function. Essentially every reference to an ontological element is no more than quantized angular momentum (or a sum of such) in each of the planes of that n-dimensional space. That says my geometric proof has to be valid. The geometric proof is nothing more than the space specification of the momentum representation laid out here. I won't go through the actual solution of quantum numbers as the math is probably well beyond anything anyone on this forum has seen but it would be quite straight forward to a decently educated theoretical physicist. In serves no real purpose as it ends up consisting of no more than another numerical representation of the defined state. Simply put, any quantum mechanical system can be specified by either a complete specification of position or a complete specification of momentum. They are essentially equivalent representations. However, there is one additional curiosity about the scheme laid out above. I have gone through developing the solution in detail and discovered a rather interesting consequence. The number of eigenvalues needed to describe a solution is exactly n. Having identified the underlying quantum numbers required to classical angular momentum quantization (neglecting spin and iso-spin which arise from the alpha operator) would be expected to attach an “l” and “m” to each ontological element reference. If our picture is to be that this universe, described by n+1 pieces of data, consists of N=(n+1)/3 three dimensional objects (or n=3N-1 eigenvalues), classical quantum would lead us to expect 2N angular momentum eigenvalues (l and m for each entity) which leads to the fact that we have N-1 eigenvalues unaccounted for (n-2N=3N-1-2N=N-1). We are thus left with exactly N-1 additional unexplained mechanical eiganvalues. If one adds one term to provide scaling for the represented undefined quantized property, one then appears to have one additional eigenvalue for each and every three dimensional event in that 3 dimensional universe. What property is it that might be quantized? Could it be mass? As I have said many times, I think what I have discovered opens some interesting doors. Doors any real scientist should be interested in looking at. Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

AnssiH Posted July 13, 2011 Report Share Posted July 13, 2011 Thanks Anssi, I appreciate your reading that thing through. I don't expect you to follow this post but rather put it forth for interests sake. It is certainly interesting. I won't make an attempt to verify the math, especially since you did not even lay it all down, I'll just follow the implications assuming your math checks out. Well, if you were to take the standard position held by the scientific community (i.e., that an explanation which explains everything known without a flaw of any kind is a correct explanation of reality) then it must be that the universe IS a rotating minimal n-dimensional polygon with unit edges. That is about the simplest explanation I have ever heard and, if you can't prove it is wrong, by Occam's principle, it certainly stands above any other “theory”. To quote Wikipedia, “in the scientific method, Occam's razor is not considered an irrefutable principle of logic, and certainly not a scientific result”, however, if one can prove it is error free, it seems to me that the issue of irrefutable logic can sort of be laid aside. Well yeah, not that anyone should take simplicity of a representation as a measure of its ontological correctedness, but I realize why you are making that comment. Well, onto the real meat; I will tell you how I was first led to the necessity of that result. If you look at my deduction of Schrödinger's equation you will notice that I divide the labels “x” into three independent sets which I call x, y and z thus leading to a three dimensional picture (the hypothetical tau is still necessary for exactly the reason it was originally necessary). This was done because Newton's equations are rather useless in a one dimensional picture; a little better in two dimensions but really valuable in three dimensions. (Perhaps the reason we think we exist in a three dimensional universe: i.e., it is the simplest picture which has any real value to our survival. Yeah, perhaps. I was thinking earlier about that issue, and I thought that a 2D representation of the same information that is currently represented in 3D form in my head, would require a radically larger number of defined objects and their associated context-sensitive rules. And even more so in 1-dimensional form. I.e. it is possible to contain more of that information into a single defined entity when that entity is associated with more dimensions. And on the other hand, there's probably a break-even point somewhere, when adding more dimensions to the representations - while makes it possible to express the circumstances in terms of less objects - it starts making the behaviour of the individual objects more complex, and that makes the representation more complex again. Or maybe three dimensional form is just "enough" for survival, and more would be a wasted effort. But I can't analytically prove any of this, it's just thoughts I'm having. Anyway, onto your thoughts about the issue; What immediately occurs to any thinking person at that point is, why not look at a four dimensional representation of reality. The problem with a four dimensional representation is that it is, as far as I am aware, impossible to picture in our minds eye (remember, it's five dimensional after we add in that tau). Thus problem must be examined from an analytical perspective and it is not easy to express how it would appear to the proper sensing organ. If we have to do the problem analytically, why stop at four. It is entirely possible that some higher dimensional representation could yield a more rational picture. In fact, why not view the universe as one point in an n dimensional Euclidean space: that is, if we have n pieces of data, why not an n dimensional model. One immediate advantage of using an n dimensional model to represent n numbers in the observation is that we no longer need to deal with the problem of two knowable events falling on the same point; note however that we still need a background of unknowable data to constrain the knowable event to actual observation. Clearly, since we have only one knowable point and our fundamental equation is valid only in the center of mass system, the probability function which represents our event must be symmetric about the origin (or we can move the origin to make the statement true). The unknowable data can be viewable as forming a potential well which constrains our point to the origin: by simple symmetry it cannot depend on any angles. The interesting thing about this model is that the radial function, the only part where any variation is possible, is totally unimportant. Clearly, any solution for the entire universe must be dimensionally scalable and we only have one linear dimension: i.e., "r". In this view, the fundamental equation of the universe (seen as a point in an n+1 dimensional sphere, n angles and one radius) becomes Yup, I believe I am following all of that. [math]\left\{\vec{\alpha}\cdot\vec{\nabla}+\beta g®\right\}\vec{\Psi}=k\frac{\partial}{\partial t}\vec{\Psi}=ikm\vec{\Psi}[/math] where we have explicitly inserted conservation of energy. Consistent with previous practice, we may use the implied operator identity [math]\vec{\alpha}\cdot\vec{\nabla}+\beta g®=ikm[/math] to generate the n-dimensional Laplacian equation [math]\left\{\nabla^2+2g^2®\right\}\vec{\Psi}=-2k^2m^2\vec{\Psi}=-K\vec{\Psi}[/math] Note the two arises because of the fact that [math]\alpha^2=\frac{1}{2}[/math]. The curious thing about the equation above is that the angular part, the only significant part, admits of solution in closed form. Note that the (n+1) dimensional Dalembertian operator (that would be n angles and one radial measure) may be written [math]\nabla^2=\sum_{i=0}^n\frac{\partial^2}{\partial x^2}= \left(\frac{1}{r}\right)^2\frac{\partial}{\partial r}r^n\frac{\partial}{\partial r}+\frac{1}{r^2}\sum_{i=1}^n\left(\prod_{l=1}^{i-1}csc^2\theta_l \right)(csc\theta_i)^{n-i}\frac{\partial}{\partial \theta_i}(sin\theta_i)^{n-i}\frac{\partial}{\partial \theta_i} [/math] essentially written in hyper spherical coordinates where [math]r=\sqrt{\sum_{i=0}^n x_i^2}[/math] [math]\theta_{i+1}= arccos\left(\frac{x_i}{\sqrt{\sum_{j=i}^n x_j^2}}\right)[/math] and [math]x_i=rcos\theta_{i+1}\prod_{l=1}^i sin \theta_l[/math]. Note that theta sub n is defined to be zero otherwise one ends up defining an [math]x_{n+1}[/math], a slight inconsistency since our n+1 arguments include “r”. Substitution (I don't expect you to confirm this as the math is not trivial) will confirm that the fundamental equation in this form has a general solution of the form [math]\Psi=\frac{1}{\sqrt{r^n}}U®\prod_{i=1}^n\Phi_i(\theta_i)[/math] Yup, I believe I am taking the validity of that on faith right now :) The result sounds quite reasonable though; which is essentially the form of an n dimensional angular momentum function. Essentially every reference to an ontological element is no more than quantized angular momentum (or a sum of such) in each of the planes of that n-dimensional space. That says my geometric proof has to be valid. The geometric proof is nothing more than the space specification of the momentum representation laid out here. Yeah, I see. It's not very long leap from seeing some set of information represented in terms of spatial coordinates of an n-dimensional entity, to seeing the same information embedded in angular momentum of an n-dimensional entity. And I can see why that would make you think of the n-dimensional polyhedron type of proof. I won't go through the actual solution of quantum numbers as the math is probably well beyond anything anyone on this forum has seen but it would be quite straight forward to a decently educated theoretical physicist. In serves no real purpose as it ends up consisting of no more than another numerical representation of the defined state. Simply put, any quantum mechanical system can be specified by either a complete specification of position or a complete specification of momentum. They are essentially equivalent representations. Well if you think it might have any value, you could still just lay the math down for easy reference of any educated physicist who might stumble across this post in the near or distant future. However, there is one additional curiosity about the scheme laid out above. I have gone through developing the solution in detail and discovered a rather interesting consequence. The number of eigenvalues needed to describe a solution is exactly n. Having identified the underlying quantum numbers required to classical angular momentum quantization (neglecting spin and iso-spin which arise from the alpha operator) would be expected to attach an “l” and “m” to each ontological element reference. What do "l" and "m" stand for here? If our picture is to be that this universe, described by n+1 pieces of data, Why "n+1 pieces of data"? consists of N=(n+1)/3 three dimensional objects (or n=3N-1 eigenvalues), classical quantum would lead us to expect 2N angular momentum eigenvalues (l and m for each entity) which leads to the fact that we have N-1 eigenvalues unaccounted for (n-2N=3N-1-2N=N-1). We are thus left with exactly N-1 additional unexplained mechanical eiganvalues. If one adds one term to provide scaling for the represented undefined quantized property, one then appears to have one additional eigenvalue for each and every three dimensional event in that 3 dimensional universe. What property is it that might be quantized? Could it be mass? You are using a language that I am not very familiar with, but I believe I understand what you are saying. That your result implies, that the 3 dimensional representation of information that we are using, has one additional "momentum quantized dimension" worth of information embedded to each element. And as I well know, you are using the concept of tau-dimension with infinite uncertainty to position to represent that information, and that it just so equals exactly to the modern physics definition of mass if you make the appropriate approximations to the solutions of the fundamental equation. It's quite interesting, I really hope there are more people than just me who understand what you are saying there. As I have said many times, I think what I have discovered opens some interesting doors. Doors any real scientist should be interested in looking at. Indeed. And if I'm capable of understanding this thing, I'm sure there are plenty more out there. -Anssi Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.