Doctordick Posted August 11, 2006 Report Share Posted August 11, 2006 I am of the opinion that the following proof is of great significance when one goes to consider "emergent" phenomena and the complexity achievable from simple constructs. The proof concerns a careful examination of the projection of a trivial geometric structure on a one dimensional line element. The underlying structure will be an n dimensional rigid entity defined by a collection of n+1 points connected by lines (edges) of unit length embedded in an n dimensional Euclidean space (i.e., a minimal n dimensional equilateral polyhedron; the generalized concept of a higher dimensional equilateral triangle with unit edges). The universe (the collection of information to be analyzed) of interest will be the projection of the vertices of a that polyhedron on a one dimensional line element. The logic of the analysis will follow the standard inductive approach: i.e., prove a result for the cases n=0, 1, 2 and 3. Thereafter prove that if the description of the consequence is true for n-1 dimensions, it is also true for n dimensions. The result bears very strongly on the range of complexity of "emergent" phenomena given an extremely simple source. First of all, the projection will consist of a collection of points (one for each vertex of that polyhedron) on the line segment of interest. Since motion of that polyhedron parallel to the given line segment is no more than uniform movement of every projected point, we can define the projection of the center of the polyhedron to be the center of the line segment: i.e., linear motion of the polyhedron has no real consequences. Furthermore, as the projection will be orthogonal to that line segment and the n dimensional space is Euclidean, any motion orthogonal to that line segment introduces no change in the projection whatsoever. It follows that the only motion of the polyhedron which provides any interesting changes in the distribution of points on the line segment will be rotations of the polyhedron in the n dimensional space. 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. Before we proceed to the proof, one issue of significance must be brought up. That issue concerns the scalability of the distribution. I referred to the collection of points on the line segment as the "universe of interest" as I want the student to think of that distribution of points as a universe: i.e., any definition of length must be arrived at via some defined characteristic of the the distribution itself or some subset of the distribution. Thus any two distributions which differ only by a scale factor will be considered to be identical distributions. Case n=0 is trivial as the polyhedron consists of one point (with no edges) and resides in a zero dimensional space. It's projection on the line segment is but one point (which, from the above constraints, is at the center of the line segment by definition) and no variations in the distribution of any kind are possible. Neither is it possible to define length within that "universe". It follows trivially that every conceivable distribution of a lone point on a line segment where the center of the distribution is defined to be the center of the line segment is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=0 (or at least can be interpreted in a way which makes it valid). I said it was trivial; it is only here for continuity in that it lets me begin with one point. Case n=1 is also trivial as the polyhedron consists of two points and one edge residing in a one dimensional space. Since the edge is to have unit length, one point must be a half unit from the center of the polyhedron and the other must be a half unit from the center in the opposite direction. Since rotation is defined as the trigonometric conversion of one axis of reference into another, rotation can not exist in a one dimensional space. It follows that our projection will consist of two points on our line segment. We can now define both a center (defined as the midpoint between the two points) and a length (define it to be the distance between the two points) in this universe but there is utterly no use for our length definition because there are no other lengths to measure. It follows trivially that every conceivable distribution of two points on a line segment (which is one) is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=1. Case n=2 is the first case which is not utterly trivial. Fabrication of an equilateral n dimensional polyhedron is not (in general) a trivial endeavorer. In order to keep our life simple, let us construct our equilateral polyhedron in such a manner so as to make the initial orientation of the lower order polyhedron orthogonal to the added dimension. Thus we can move the lower order entity up from the center of our new coordinate axis and add a new point on the new axis below the center. In this case, the coordinates of previous polyhedron (as displayed in the n Euclidean space) remain exactly what they were for the n-1 established coordinates and are all shifted by the same distance from zero along the new axis. The new point has a position zero in all the old coordinates (it is on the new axis) and an easily calculated position on in the negative direction on that new axis (that distance must be equal to the new radius of the vertices of the old polyhedron as measured in the new n dimensional space). The proper movement is quite easy to calculate. Consider a plane through the new axis and a line through any vertex on the lower order polyhedron. If we call the new axis the x axis and the line through the chosen vertex the y axis, the y position of that vertex will be the old radius of the vertex in the old polyhedron. 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 radius is exactly the same distance the new point must be from the new center: i.e., its radial distance. Assuring the new edge length will be unity imposes a second Pythagorean constraint consisting of the fact that the old radius squared plus (the new radius plus the distance the old polyhedron was moved up) squared must be unity. [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] 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] The case n=0 was a single point in a zero dimensional space. The case n=1 can be seen as an addition of one dimension x_1 (orthogonal to nothing) where point #1 was moved up one half unit in the new dimension and a point #2 was added at minus one half in the new dimension (both the new radius and "distance to be moved up" are one half). The case n=2 changes the radius to one over the square root of three and the line segment (the result of case n=1) must be moved up exactly one half that amount. A little geometry should convince you that the result is exactly an equilateral triangle with a unit edge length. Projection of this entity upon a line segment yields three points and the relative positions of the three points are changed by rotation of that triangle. In this case, we have two points to use as a length reference and a third point who's distance from the center can be specified in terms of that defined length reference. Using those definitions, two of the points can be defined to be one unit apart and the third point's position can vary from any specific position from plus infinity to minus infinity. The infinities are approached when the edge defined by the two vertices being used as our length reference approaches orthogonality to the line segment upon which the triangle is being projected (in which case the defining unit of measure falls towards zero). Plus infinity would be when the third point is on the right (by convention) and minus infinity when the third point is on the left (by common convention, right is usually taken to be positive and left to be negative). It thus follows that every conceivable distribution of three points on a line segment is achievable by a particular rotational orientation of the polyhedron (our triangle). Thus the theorem is valid for n=2. Case n=3 consists of a three dimensional equilateral polyhedron consisting of four points, six unit edges and four triangle faces: i.e., what is commonly called a tetrahedron. If you wish you may show that the radius of vertices is given by one half the square root of three halves and the altitude by the radius plus one over two times the square root of six (as per the equations given above). In examining the consequences of rotation, to make life easy, begin by considering a configuration where a line between the center of our tetrahedron and one vertex is parallel to the axis of projection on our reference line segment. Any and all rotations around that axis will leave that vertex at the center of our line segment and actually consist of rotation in the plane of the face opposite to that point. Essentially, except for that particular point, we obtain exactly the same results which were obtained in case n=2 (that would be projection of the triangle face opposite the chosen vertex). Using two of the points on that face to specify length, we can find an orientation which will yield the third point in any position from minus infinity to plus infinity while the forth point remains at the center of the reference line segment. Having performed that rotation, we can rotate the tetrahedron around an axis orthogonal to the first rotational axis and orthogonal to the line on which the projection is being made. This rotation will end up doing nothing to the projection of the first three points except to uniformly scale their distance from the center. Since we have defined length in terms of two of those points, the referenced configuration obtained from the first rotation does not change at all. On the other hand, the forth point (which was projected to the center point) will move from the center towards plus or minus infinity depending on the rotation direction (the infinite positions will correspond to the orientation where the line of projection lies in that face opposite the fourth point: i.e., the scaled reference distance approaches zero). It follows that all possible configurations of the four points in our projection can be reached via rotations of the tetrahedron and the theorem is valid for n=3. The final part of the proof (if it is true for an n-1 dimensional figure, it is true for an n dimensional figure) requires a little thought: Since the space in which the n dimensional polyhedron is embedded is Euclidean, we can specify a particular orientation of that polyhedron by listing the n coordinates of each vertex. That coordinate system may have any orientation with respect to the orientation of the polyhedron. That being the case, we are free to set our coordinate system to have one axis (we can call it the x axis) parallel to the line on which the projection is to be made. In that case, except for a scale factor (which must be obtained from the distribution), a list of the x coordinates of each point correspond exactly to the apparent positions of the projected points on our reference line. Thus I will henceforth use the x axis in the n dimensional space as a surrogate for my reference line segment. If the theorem is true for an n-1 dimensional polyhedron, there exists an orientation of that polyhedron which will correspond to any specific distribution of n points on a line (where scale is established via some procedure internal to that distribution of points). If that is the case, we can add another axis orthogonal to all n-1 axes already established, move that polyhedron up along that new axis a distance equal to [math]x_n = \frac{1}{\sqrt{2n(n+1)}}[/math] and add a new point at zero for every coordinate axis except the nth axis where the coordinate will be set to [math]-r_n = x_n = - \sqrt{\frac{n}{2(n+1)}}[/math]. The result will be an n dimensional equilateral polyhedron with unit edge which will project to exactly the same distribution of points obtained from the previous n-1 dimensional polyhedron with one additional point at the center of our reference line segment. If our n dimensional polyhedron is rotated on an axis perpendicular to both the reference line segment and the nth axis just added, the only effect on the original distribution will be to adjust the scale of every point via the scale factor [math]cos\theta[/math]. That is, the new [math]x_i[/math] is obtained by [math]x_{1i New} = x_{1iOld}cos\theta + \sqrt{\frac{n}{2(n+1)}}sin\theta [/math], where theta is the angle of rotation (notice that the sin term yields a simple shift exactly the same for all points which is quite meaningless as far as the pattern of those points is concerned). Meanwhile, the [math]x_1[/math] position of the added point will be given exactly by [math]-r_n sin \theta[/math] (the cos term vanishes as it started on the origin of [math]x_1[/math]). Once again, the added point may be moved to any position between plus and minus infinity which occur at plus and minus ninety degrees of rotation. Once again, the length scale is to be established via some procedure internal to the distribution of points. It follows that the theorem is valid for all possible n. QED 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. We can go one step further. Within those n-2 axes orthogonal to the x and y axes, one can choose one to be the z axis and still have n-3 definable planes orthogonal to both the x and the y axes. That provides one with n-3 possible rotations which will leave the projections on the x and y axes unchanged. Since, in the construction of our polyhedron no consequences of rotation had any effect until we got to rotations after addition of the third point, these n-3 possible rotations are sufficient to obtain any distribution of projected points on the z axis without altering the established projections on the x and y axes. Thus it is seen that absolutely any three dimensional universe consisting of n+1 points for n greater than four can be seen as an n dimensional equilateral polyhedron with unit edges projected on a three dimensional space. That any means absolutely any configuration of points conceivable. Talk about "emergent" phenomena, this picture is totally open ended. Any collection of points can be so represented! Consider the republican convention at noon of the second day (together with every object and every person in the rest of the world; and all the planets; and all the galaxies ...) where the collection of the positions of all the fundamental particles in the universe is no more than a projection of some n dimensional equilateral polyhedron of unit size on a three dimensional space. Talk about emergent phenomena! On top of that, if nothing in the universe can move instantaneously from one position to another, it follows that the future (another distribution of that collection of positions of all the fundamental particles in the universe) is no more than another orientation of that n dimensional polyhedron and the evolution of the universe in every detail must correspond to continuous rotation of that figure. Think about that view of a rather simple geometric construct and the complex phenomena which is directly emergent from the fundamental perspective. (And every variable associate with any motion is quantized; it is all angular motion!) Have fun – Dick Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 11, 2006 Report Share Posted August 11, 2006 I am of the opinion that the following proof is of great significance when one goes to consider "emergent" phenomena and the complexity achievable from simple constructs. The proof concerns a careful examination of the projection of a trivial geometric structure on a one dimensional line element. First of all, the projection will consist of a collection of points (one for each vertex of that polyhedron) on the line segment of interest. 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. This sounds much like Plato's divided line, but with speciation of the divisions. On top of that, if nothing in the universe can move instantaneously from one position to another, it follows that the future (another distribution of that collection of positions of all the fundamental particles in the universe) is no more than another orientation of that n dimensional polyhedron and the evolution of the universe in every detail must correspond to continuous rotation of that figure. Think about that view of a rather simple geometric construct and the complex phenomena which is directly emergent from the fundamental perspective. Have fun – DickHave you read Buckminster Fuller's Synergetics Doc? He rather takes a similar view, albeit he uses far more words, symbols, and geometric figures. :shrug: Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 12, 2006 Report Share Posted August 12, 2006 This sounds much like Plato's divided line, but with speciation of the divisions.Here is the passage I referred to: The Republic, page 252 in Chapter 6:Plato said:Now take a line which has been cut into two unequal parts, and divide each of them again in the same proportion, and suppose the two main divisions to answer, one to the visible and the other to the intelligible, and then compare the subdivisions in respect of their clearness and want of clearness, and you will find that the first section in the sphere of the visible consists of images. And by images I mean, in the first place, shadows, and in the second place, reflections in water and in solid, smooth and polished bodies and the like: Do you understand? http://www.literaturepage.com/read/therepublic-252.html Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 16, 2006 Author Report Share Posted August 16, 2006 Hi Turtle, I hate to say it but I was really disappointed with your posts. I sincerely wonder if you understood the proof and what it implies. Please tell me sincerely what you think the implications of that proof are. Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 16, 2006 Report Share Posted August 16, 2006 Hi Turtle, I hate to say it but I was really disappointed with your posts. I sincerely wonder if you understood the proof and what it implies. Please tell me sincerely what you think the implications of that proof are. Have fun -- Dick Nah...you don't mind saying it.:eek2: I sincerely wonder if you understand the connection I attempted to draw with Fuller's rigorous descriptions of the tetrahedron & its possible orientations? Have you read Fuller's descriptions or not? If not, how do you know they don't apply? If you have read it, then how is it relevant or not? You say you have a geometric proof, & yet I see no geometric figures; no doubt the figures don't apply any more than Venn diagrams do to your sets.:) You seem to imply that the patterns on the line may look the same with different orientations of the tetrahedron.:eek2: Where is your line & how oriented to which tetrahedron in this Fuller graphic?http://www.rwgrayprojects.com/synergetics/s01/figs/f00103.html:eek2: Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 16, 2006 Author Report Share Posted August 16, 2006 Nah...you don't mind saying it.:eek: I sincerely wonder if you understand the connection I attempted to draw with Fuller's rigorous descriptions of the tetrahedron & its possible orientations?No, I have no clue as to what connection you had in mind as the URL you gave doesn't yield anything to my browser except what appears to be the book cover. The title provides a link to what appears to be a discussion of issues concerning the web site which implies the text is somewhere but I couldn't find it anywhere even when I googled "synergetics". You say you have a geometric proof, & yet I see no geometric figuresI expected you to conceive of the relevant figures in your minds eye as I have no idea as to how to draw an n dimensional solid; even one so simple as to be the simplest conceivable. In this case, n's of interest certainly exceed a billion and the best example would be for n to be the number of electrons plus the number of protons plus the number of neutrons in the physical universe. Their approximate distribution pattern would give one a pretty good picture of the physical universe. The proof is a simple analytical geometry proof that such a distribution is exactly representable by a three dimensional projection of an n dimensional object entirely analogous to the n dimensional generalization of a rigid tetrahedron. Since every possible set distributions is so representable, the idea of continuous existence of those elements implies the temporal evolution of the universe is representable by a smooth rotation of that same figure.I can however ; no doubt the figures don't apply any more than Venn diagrams do to your sets.:evil:This comment implies that you totally misunderstood that earlier discussion I had with someone concerning Venn diagrams. Venn diagrams apply as well to my sets as they do to any other sets. What doesn't apply is the issue Venn diagrams are used to represent. Venn diagrams are used to represent intersections, combinations and exclusion relationships between different sets of the same items. My sets A, B, C and D have no possible "intersection", "combinations" or "exclusion" relationships of serious significance between them as they are, each one except B, sets of quite different entities. A is defined to be "that which is to be explained". Since B(t) consists of a change in what you think you know of A it is presumed to have some intersection with A but it also contains elements not in A; however, since what is and is not part of A is strictly impossible to differentiate (if you can differentiate between them, your explanation is patently wrong), the issues presented by Venn diagrams are pretty worthless. The sets C and D are specifically defined to have no intersections and finally the "elements" of C and D are totally different cuts of specific sets B(t) so intersections once again become a worthless issue. My point with whoever it was, was that Venn diagrams have nothing to do with the relationships of interest developed in that presentation. It really makes little difference as I doubt anyone understood what I was saying anyway.You seem to imply that the patterns on the line may look the same with different orientations of the tetrahedron.:D Looking the same wasn't the real issue. What the proof says is that for every possible projection pattern of n-1 of the n vertices of that n-1 dimensional "entrahedron" (I just made up that name, n – trahedron ;) ) there exists a rotation which leaves that pattern unchanged (except for scaling) while moving the nth vertex through every possible position relative to that pattern. The fact that the proof is true means that there exists a specific rotation which will move any point to any position while leaving all the others unchanged (except for scaling). Since scaling, any definition of physical "length", must be defined in terms of sub patterns within the pattern, it follows that there exists no pattern which is not perfectly representable by a specific orientation of that "entrahedron".Where is your line & how oriented to which tetrahedron in this Fuller graphic?http://www.rwgrayprojects.com/synergetics/s01/figs/f00103.html:hyper:It isn't the line's orientation which is significant here; it is the rotations which are significant. Put the line of interest anywhere and develop the positions of the projections of the four vertices. Look at planes orthogonal to that line containing the vertex of interest (that will define four unique planes). The four points where those four planes intersect that line provide a specific pattern. Now examine how that pattern changes as you rotate the tetrahedron around a line through the center of the tetrahedron and one vertex of the tetrahedron. You should notice that the projection of the vertex you have chosen does not move. On the other hand, the three other points (on the face opposite that vertex) will move through all possible patterns (except for scale) achievable with three points on a line. Now, having achieved whatever pattern you chose for the projection of those three vertices, establish a plane through the center of that face (the face opposite the vertex you chose for the earlier rotation) orthogonal to the reference line. That plane intersects the center of the face of interest and provides us a line of intersection orthogonal to the reference line and orthogonal to the axis of rotation above (every line in that face is orthogonal to the first rotation axis). If you rotate the tetrahedron around that line, the pattern you achieved in the first rotation will change in scale only while the vertex you started with will move through all possible positions achievable (except for scaling). Thus it is clear that any pattern of four points can be achieved (except for scaling). What I have proved is this analysis carries all the way to the number of physical entities in the universe (and can be carried up to three orthogonal projections: i.e., projection into a three dimensional space). It follows that any possible physical distribution of three dimensional entities making up the universe can be seen as a three dimensional projection of a rigid unit "entrahedron". And the temporal evolution of the universe can be seen as smooth rotation of that "entrahedron". I hope that is a little clearer. :eek: Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 16, 2006 Report Share Posted August 16, 2006 No, I have no clue as to what connection you had in mind as the URL you gave doesn't yield anything to my browser except what appears to be the book cover. The title provides a link to what appears to be a discussion of issues concerning the web site which implies the text is somewhere but I couldn't find it anywhere even when I googled "synergetics". Here is the table of contents the title page leads to:http://www.rwgrayprojects.com/synergetics/toc/toc.html This comment implies that you totally misunderstood that earlier discussion I had with someone concerning Venn diagrams. Venn diagrams apply as well to my sets as they do to any other sets. That someone is me, although perhaps that's an irrelevant detail. Back then you told me explicitly Venn diagrams didn't apply. :hyper: Now you say they do but conditionally. Now examine how that pattern changes as you rotate the tetrahedron around a line through the center of the tetrahedron and one vertex of the tetrahedron. Have fun -- DickFrom this diagram, which halving of the tetrahedron do you refer to?http://www.rwgrayprojects.com/synergetics/s01/figs/f00103.html This may be an ironic statement in the sense you are trying to explain explanations, but I can only go from what I think I know about geometry in my attempt to understand your explanation of what you think I don't know about geometry.:eek2: Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 17, 2006 Author Report Share Posted August 17, 2006 That someone is me, although perhaps that's an irrelevant detail. Back then you told me explicitly Venn diagrams didn't apply. :) Now you say they do but conditionally.I think we are just failing to communicate. Venn diagrams are simply of no use in understanding the issues under discussion at that time: i.e., none of the serious issues I was presenting could be clarified through Venn diagrams; no more than an understanding of photographic exposure analysis would help one see my arguments.From this diagram, which halving of the tetrahedron do you refer to?Where do you get the idea I am, at any time, referring to "halving" of a tetrahedron. I am talking about projection of the vertices of a spacial rotating tetrahedron onto an arbitrary line (and, by the way, that issue is only of interest in the case n=3). The case n=0 is projection of a point onto an arbitrary line; case n=1 is projection of a the end points of a line onto an arbitrary line and case n=2 is projection of an equilateral triangle onto an arbitrary line. The case n=2 is the first time rotation has any meaning; that is why the inductive argument should include the case n=3 (otherwise, we are talking about new phenomena, not expressible in a lower dimension: i.e., "rotation" is fundamentally a meaningless concept in both a zero dimensional space and a one dimensional space). This may be an ironic statement in the sense you are trying to explain explanations, but I can only go from what I think I know about geometry in my attempt to understand your explanation of what you think I don't know about geometry.:doh:This thread has nothing to do with explaining explanations; I am simply putting forth a simple geometric proof. I had presumed you knew enough about Euclidean geometry to follow the proof. At the moment, I am at a complete loss as to what problem you are having. I could find nothing in Fuller's "Synergetics" which had anything to do with my geometric proof or shed any light on your difficulties in understanding that proof. I am totally in the dark and can not seem to make any sense of your responses at all. They don't seem to have any bearing on the geometric proof that I can see. Sorry about that, I thought geometry was your forte -- Dick Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 17, 2006 Report Share Posted August 17, 2006 I think we are just failing to communicate. ... I concur.Where do you get the idea I am, at any time, referring to "halving" of a tetrahedron. I am talking about projection of the vertices of a spacial rotating tetrahedron onto an arbitrary line (and, by the way, that issue is only of interest in the case n=3). ...The diagram shows the 2 ways you can rotate a tetrahedron; I was simply asking which type of rotation you refer to.Sorry about that, I thought geometry was your forte -- Dick So am I to understand you can't make any drawings of your geometry? The multi-dimensional aspect is no barrier as I understand it, having seen planar representations of the complex Calibi-Yau shapes used in multidimensional string theory. I tend to understand geometry first visually & then work back to the algebraic representations. :) Quote Link to comment Share on other sites More sharing options...

ughaibu Posted August 18, 2006 Report Share Posted August 18, 2006 Does this show the "complexity achievable from simple constructs", or the simple constructs to which complexity can be reduced? Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 18, 2006 Author Report Share Posted August 18, 2006 The diagram shows the 2 ways you can rotate a tetrahedronFuller is apparently interested in specific rotations yielding symmetries. There are an infinite number of ways you can rotate a tetrahedron, not just the two he presents. What you need to do is define the axis of rotation and, in a general rotation in three dimensions, that axis can be any line you wish it to be (i.e., define the line and you define the rotation). The first rotation (which he refers to as "parallel halving") is on an axis orthogonal to two opposite edges and through the center of the tetrahedron. The second rotation (which he refers to as "Perpendicular Halving") is on an axis which is identical to a specific edge of his tetrahedron. I have no idea where you get the idea that those are the only possible ways in which the tetrahedron can be rotated.So am I to understand you can't make any drawings of your geometry?No, I wouldn't say that. What I would say is that it is easier to describe the rotations of interest than to draw them. Furthermore, drawing them on this forum presents two major problems for me: first, I do not possess any convenient software for drawing computer images and second, at the moment, I do not know the mechanism to insert a sized image file on this forum (i.e., it isn't a convenient thing for me to do without a little work).I tend to understand geometry first visually & then work back to the algebraic representations. :umbrella:Well, I have found that "visualizing things" is pretty well limited to three dimensions and if one wants to work with the details of multidimensional constructs, analytical representation is much more productive. Once you have the analytical representations, you can create planar cuts to simplify the problem of visualizing important aspects. I will see what I can do about creating a drawing of the rotations I spoke of in my previous post but don't expect a quick answer. I tried doing the case n=3 with windows "paint" and found it very difficult to cram all the required surfaces in without manufacturing a totally cramped image. There are just too many things which have to be represented: the tetrahedron itself, the line on which the projections are to be made, the axis of rotation and then five planes which provide the projections (four for the four vertices of the tetrahedron and one for the center of the face opposite the vertex of the first rotation). Personally, I think the thing is much easier to do analytically but I will see what I can do. :shrug: Does this show the "complexity achievable from simple constructs", or the simple constructs to which complexity can be reduced?I would say those are little more than two sides of the same coin. In essence, it shows both depending on the way you perceive the proof. :Glasses: Have fun -- Dick Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 19, 2006 Report Share Posted August 19, 2006 Well, I have found that "visualizing things" is pretty well limited to three dimensions and if one wants to work with the details of multidimensional constructs, analytical representation is much more productive. Once you have the analytical representations, you can create planar cuts to simplify the problem of visualizing important aspects.Have fun -- DickThis is presumptuous of what others' visualizations reveal. As an experienced instructor, one has to accomodate many learning styles such as visual, auditory, repetition etc.. It is also incongruent to title the thread "simple geometric proof" & not offer a simple geometric diagram. On an operational note, if you have access to a scannner, then just sketch a simple diagram on a scrap of paper, scan it, & post it as an attachment to this thread. Post Script: The essence of a coin is not in its two sides, but in its one edge. - Roger thelonoius George Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted August 21, 2006 Report Share Posted August 21, 2006 Turtle, the difficulty is mainly a matter of generalizing to an arbitraru number n of dimensions. It isn't really hard to understand for up to a tetrahedron, except that a clearer explanation might help. I've got to go soon, after a looooooong talk with the guy I'm currently doing my job for (which is in n dimensions but a lot more complicated than the above theorem), but try to think of a tetrahedron that you can move wrt a straight line and also enlarge or shrink in exact scale. Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 21, 2006 Report Share Posted August 21, 2006 Turtle, the difficulty is mainly a matter of generalizing to an arbitraru number n of dimensions. It isn't really hard to understand for up to a tetrahedron, except that a clearer explanation might help. I've got to go soon, after a looooooong talk with the guy I'm currently doing my job for (which is in n dimensions but a lot more complicated than the above theorem), but try to think of a tetrahedron that you can move wrt a straight line and also enlarge or shrink in exact scale. No drawing from you either Q? Not even a simple case? Is wrt a typo or an abbreviation? Honestly, if this is so simple to understand, why is it so hard to explain?:) Quote Link to comment Share on other sites More sharing options...

Doctordick Posted August 21, 2006 Author Report Share Posted August 21, 2006 Hi Turtle, I think we are having great difficulty communicating. We both seem to be misinterpreting the comments of the other. As I am sure Qfwfq will attest, I am not the best communicator in the world. From his "clearer explanation" I get the feeling he doesn't follow the proof either (but I am probably wrong). The acronym "wrt" could mean "with relation to" but I am not too good at acronyms so I don't presume to know what is intended (could be a misspelling of "with"). Back before I retired, I was on a corporate business trip on a small private jet and several of the young whipper-snappers (if you have ever heard that term, it comes from snapping whips for attention back when we used whips regularly) were talking about the complexity of their schedules at our destination. One of them mentioned several times that he had to make an important "BM" as soon as we got there. From the conversation, I soon got the impression he wasn't talking about a "bowel movement" so I finally asked him what he meant by a "BM"; he said a "BM" was a business meeting. I guess it's just hard to teach an old dog new tricks. Well, I went to the trouble of drawing the case n=3 diagram. The initial drawing was so cluttered (with the five projection planes) that it was almost impossible to see which plane was where so I altered the shapes of four of the planes so the eye could follow where they went. I think that the only one that is difficult to see from the diagram is the one that provides the second axis of rotation. Since that one provides a "line of intersection" with the top face of the tetrahedron, it really can't be cut down. By the way, I put the dotted line on the plane going through vertex #4 because the bottoms of all the other planes terminate at the reference line and I wanted it to be obvious that vertex #4 is by far the lowest point on the figure. Now visualize the tetrahedron rotating around the first axis. That rotation will move neither the center point nor vertex #4 on the reference line but the positions of the other three points will change. Rotate the tetrahedron on the second axis and the center point is the only point which will not move. Vertices #1, #2 and #3 will move but will maintain exactly the same pattern achieved in the first rotation (except for scale). Vertex #4 will move through all positions on the reference line which (when adjusted for the scale changes required to maintain the pattern of Vertices #1, #2 and #3) will include every point from plus to minus infinity. The infinities occur when the top face of the tetrahedron is rotated into the projection plane at which time the reference used as a standard distance (no matter how it was achieved) will go to zero on the reference line. This is presumptuous of what others' visualizations reveal.Not really; a decent handle on analytic geometry is absolutely essential if one wants to analyze figures in higher dimensions. I still hold that the human mind simply cannot comprehend the full complexity of higher dimensional spaces. It can only visualize aspects which represent phenomena representable in three dimensions. I hold it as quite evident that we live in a three dimensional universe for one very simple reason: because it's the simple limit imposed upon us by the phenomenal success of three dimensional mechanics at providing us with the information necessary for survival against the macroscopic aspects of our environment. I think it might be worthwhile to have a thread on the issues inherent in n dimensional Euclidean geometry. It gets complex rather quickly. By the way, I said the proof was simple, not that the inherent issues being represented were. In fact, the representation problems become impossibly complex very rapidly; yet the proof is quite simple when looked at via analytical geometry. Have fun -- Dick Turtle 1 Quote Link to comment Share on other sites More sharing options...

Turtle Posted August 22, 2006 Report Share Posted August 22, 2006 Hi Turtle, I think we are having great difficulty communicating. We both seem to be misinterpreting the comments of the other. ... Well, I went to the trouble of drawing the case n=3 diagram. The initial drawing was so cluttered (with the five projection planes) that it was almost impossible to see which plane was where so I altered the shapes of four of the planes so the eye could follow where they went. I think that the only one that is difficult to see from the diagram is the one that provides the second axis of rotation. Since that one provides a "line of intersection" with the top face of the tetrahedron, it really can't be cut down. Have fun -- Dick Absolutely what I had in mind generally, and what I did not have in mind specifically! Much appreciated Doc; thank you for that extra effort. :hihi: I have re-read the original proof with your drawing in view, and a few more dozen similar reviews may suffice to fix it all in my mind. One question comes to mind just now and that is does it ultimately matter in your planned application of this geometry to your explanation, whether the rotation is clockwise or counter? That is to say does chirality play a role?Thanks again Doc....cogitating...:shrug: + :ip: + :hyper: = :) Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted August 22, 2006 Report Share Posted August 22, 2006 WRT is With Respect To. No drawing from you either Q? Not even a simple case?Sorry, but as I said I was very short on time... Honestly, if this is so simple to understand, why is it so hard to explain?Not so simple to understand, but mainly a matter of habit. I think your difficulty is probably because it's an argument in an arbitrary number of dimensions. I would definitely say that chirality isn't the issue here. Now, looking at Dick's drawing, you see the planes at right angles to the line and these give the projection of each point. For a given pair of points, the segment of length l is projected onto one of length [math]\norm l\cos\alpha[/math] (angle formed by the segment with the planes). Thus it is trivial to scale up a given set of points on the line, by scaling up the tetrahedron, and even more trivial to perform a translation. This means that you can take two of the points on the line as reference. To find a tetrahedron that will give the remaining points (on the line), first recall that as well as rotating it around the corresponding two ti is possible to move one of these in its plane with a consequent scaling of the whole tetrahedron. For n + 1 points in n dimensions, each of the planes becomes an n - 1 dimensional space orthogonal to the line, but apart from this analogous considerations apply. A more exact argument needs to be recursive, as the one in Dick's first post. Turtle 1 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.