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| Why Not? |
Posted: Sep 6 2007, 07:26 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 327 Joined: 29-November 05 Positive Feedback: 53.33% Feedback Score: 1 |
This tread is to discuss opinions of, and possible geometries behind, structured spacetime. There is potentially a lot of ground that can be covered so I will start with my “circles”…
When first introduced to the idea of “structured spacetime”, I began to wonder if patterns could be found in geometry that would lead to revelations in physics. So beginning with the idea that spherical propagation would be the simplest to examine, I set out to model spherical propagation on a hexagonically packed lattice of unit spheres. But I soon found that two dimensional modeling was much easier with the tools I had available, AutoCAD, and so I was able to create the following “pictures” with circles instead of spheres… The Pattern: The Whole Picture: Close Up of the center: Here’s the interesting part… The background is a lattice of hexagonically packed unit circles. The “center” unit circle is arbitrarily selected. From the center circle, circumferential circles are drawn, each two times the unit circle diameter greater than the next. By constructing circumferential circles is such a manner, there are a minimum of six unit circles in each successive hexagonal packing that are internally tangential to the circumferential circle associated with that packing. So, from the center circle (packing q=0) the first packing (q=1) contains six circles that are internally tangential to a circumferential circle that has a radius of three (unit circle has a radius of one). The second packing (q=2) contains six circles that are internally tangential to a circumferential circle that has a radius of five. In the “close up of the center” drawing linked above, you can see that these six unit circles at every packing level are at 0, 1/3pi, 2/3pi, pi, 4/3pi, and 5/3pi (0, 60, 120,180, 240, 300 degrees respectively). Since the centers of these circles form the vertices of the hexagon representative of the associated packing, I have elected not to identify them in any special way. Here is the really interesting part… Whenever the hexagonally packed level equals a prime number of the form 6n+1, there are 12 additional circles that are internally tangential to the associated concentric circles. In the drawings linked above, these are represented by green shading. Additionally, all multiples of that prime also include 12 additional internally tangential unit circles. These multiple occurrences are represented by magenta shading. For example, at q=7, there are twelve additional internally tangential circles, shaded green and at q=14, 21, 28… the unit circles are shaded magenta. Notice that only when 6n+1 is prime are additional unit circles tangential to the associated concentric circle. So 7, 13, 19, 31… have additional circles but not 25, 49 (except as a multiple of 7), 55, 85… What’s more, when the packing level equals the multiple of two non identical primes greater than 5, there are 36 additional unit circles that are internally tangential to the associated concentric circle. These are represented in the drawing by cyan shading and are shown at 91 and 133. The drawings referenced are taken out to only 144 packing level. The self similar pattern in each 30 degree segment is obvious when looking at the “pattern” picture. The wave pattern is obvious when looking at the “whole” picture. What I have not been able to discover is why internally tangential unit circles only appear at primes (and multiples thereof) of the form 6n+1. Why does the same effect not appear with primes of the form 6n-1? Comments, discussion and help with the maths are most welcome. Mahalo. P.s. Tracing a curve, beginning at the center and moving clockwise (or anti-clockwise) form one green circle to the next successive green circle but in the next 60 degree segment produces are curve pretty close to an Golden Spiral. This post has been edited by Why Not? on Sep 6 2007, 07:33 PM -------------------- Nothing is certain.
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| jal |
Posted: Sep 6 2007, 07:41 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Great!
I'm going to take time to answer. Hope some other readers have already got contributing inputs in ther files, for you. Jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
| N O M |
Posted: Sep 7 2007, 02:41 AM
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on holiday, get your abuse elsewhere ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 3691 Joined: 4-December 06 Positive Feedback: 56.82% Feedback Score: 93 |
great... not -------------------- Proud owner of negative feedback from: 555Joshua, alokmohan, bee, BigFairy, Bi shadi, Bloy, Bryn Richards, bukh, Confused2, DavidD, deadbeat, Derek1148, eyeque, Farsight, fivedoughnut, freethis, Gizmo, Gorgeous, howtothinklikegod, inQZtive, insight, kaneda, landon, LeTUOtter, Majkl, meBigGirl'sBlouse, Mediocre-Minded, midwestern, Mike Adams, Mirrorman, Morpheus, Mr. Robin Parsons, newton, Nick, on2thiests, oracle1, philip347, PIATLAS, PJParent001, Precursor562, Quatermass, Raphie Frank, reasonwhy, rethinker, Samantha Hildreth, A•SHEOL, Solid State Universe, Soultechs, Squeeze, SteveA2, StevenA, stundie, Sylwester Kornowski, (name removed by request), ubavontuba, vkamath, wbraxtonwilson, xtrmn8r, Zarabtul, Zephir, [please insert name here]
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| jal |
Posted: Sep 7 2007, 03:02 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Hi Why Not?
It's going to be pretty hard to get any inputs from those who do not have any idea of what we are going to try to do. Those who believe in something else can start their own discussion somewhere else. Let's start with definitions. Isotropy http://en.wikipedia.org/wiki/Isotropic Isotropic radiation has the same intensity regardless of the direction of measurement, and an isotropic field exerts the same action regardless of how the test particle is oriented. ----------- Isotropic coordinates http://en.wikipedia.org/wiki/Isotropic_coordinates In the theory of Lorentzian manifolds, spherically symmetric spacetimes admit a family of nested round spheres ------------- Spherically symmetric spacetime http://en.wikipedia.org/wiki/Spherically_symmetric_spacetime A spherically symmetric spacetime is one whose isometry group contains a subgroup which is isomorphic to the (rotation) group SO(3) and the orbits of this group are 2-dimensional spheres (2-spheres). The isometries are then interpreted as rotations and a spherically symmetric spacetime is often described as one whose metric is "invariant under rotations". The spacetime metric induces a metric on each orbit 2-sphere (and this induced metric must be a multiple of the metric of a 2-sphere). Spherical symmetry is a characteristic feature of many solutions of Einstein's field equations of general relativity, especially the Schwarzschild solution. A spherically symmetric spacetime can be characterised in another way, namely, by using the notion of Killing vector fields, which, in a very precise sense, preserve the metric. ----------- Rotation group http://en.wikipedia.org/wiki/Rotation_group In mechanics and geometry, the rotation group is the group of all rotations about the origin of 3-dimensional Euclidean space R3 under the operation of composition. By definition, a rotation about the origin is a linear transformation that preserves length of vectors and preserves orientation (i.e. handedness) of space. A length-preserving transformation which reverses orientation is called an improper rotation. Composing two rotations results in another rotation; every rotation has a unique inverse rotation; and the identity map satisfies the definition of a rotation. Owing to the above properties, the set of all rotations is a group under composition. Moreover, the rotation group has a natural manifold structure for which the group operations are smooth; so it is in fact a Lie group. The rotation group is often denoted SO(3) for reasons explained below. ------------ Spherical harmonics http://en.wikipedia.org/wiki/Spherical_harmonics In mathematics, the spherical harmonics are the angular portion of an orthogonal set of solutions to Laplace's equation represented in a system of spherical coordinates. Spherical harmonics are important in many theoretical and practical applications, particularly in the computation of atomic electron configurations, the representation of the gravitational field, geoid, and magnetic field of planetary bodies, characterization of the cosmic microwave background radiation and recognition of 3D shapes in computer graphics. -------------- Make sure to read the links for more info....especially Spherical harmonics jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
| jal |
Posted: Sep 7 2007, 04:08 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Why Not?
I like the different approach Structured Spacetime Is there a limiting geometry... Rather than trying to "see" what there is inside the "particle/wavepacket" focusing on the spacing that they must have. On a more technical level, it is being addressed by the physic community. ----------------- http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9503/9503024v2.pdf Lectures on (2+1)-Dimensional Gravity S. Carlip 18 March 1995 ----------- I have a lot of references in my summary that can be usefull to the advanced "seekers". Getting back to your CAD drawings... If you were to make a drawing using minimum separation length (assuming that planck length is the minimum) what would you end up with? ----------- In order to get 3d it is necessary to imagine a cube. There are 6 faces .... on each face there is a "particle" (I'm going to use that word for clarity) In physic they refer to those locations/positions as 6j. If you round off the corners of your cube you end up with a sphere. (a 2d sphere) If you flatten the sphere you end up with a 2d surface. (Your CAD drawings.) If you color the 6 particles and don't color the minimum spacing you should get a pattern that would be interesting. .... keep repeating the 6j. Let's see what comes out.... jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
| Wulf |
Posted: Sep 7 2007, 08:41 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 377 Joined: 16-April 07 Positive Feedback: 69.23% Feedback Score: 5 |
This thread looks promising, I'm anxious to see where this leads.
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| Why Not? |
Posted: Sep 7 2007, 09:11 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 327 Joined: 29-November 05 Positive Feedback: 53.33% Feedback Score: 1 |
Hey Jal, Thanks for the links. I do not have much time at the moment but I was hoping you could clarify what you meant by...
It seems to me that you would end up with a radial "line" each 60 degrees from the rest for each additional packing if using unit circles. Otherwise you end up with an infinite progression of seven packed circles within each packed circle of the previous set (going from small to large). I'm not sure what you are after... Also, did you see this? Quantum Criticality Found in a Simple Liquid I have not had time to look through their web site, but here is the link if you are interested... The Low Temperature Laboratory, at Royal Holloway, at the University of London But to get back to my drawing - it's geometry. Any scale can be applied. What I am trying to figure out is if there is any significance to the fact that at only certain packing levels will more than six unit circles perfectly fit inside the associated concentric circle. Why is this? More importantly, Why does it only happen on primes and multiples there of? More later… Mahalo -------------------- Nothing is certain.
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| TRoc |
Posted: Sep 7 2007, 11:37 PM
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Hi all, WN?, I assume that the circles that you "drew" are arbitrary? I mean, that you did not go out n hexagons, and then consider that the radius. I would think that some "progression" would be needed, to answer your question about the primes. To "make a circle", you could (besides the pi example above): ![]() http://en.wikipedia.org/wiki/Hexagon or approximate with "spirals": ![]() [you could look at the "Fibonacci primes", but at the expense of a continuous set of integers] staying in the same theme, you also have:
![]() Scatter plot of the first Pythagorean triples within 4500 ![]() I think that your best chance is here: http://www.numberspiral.com/index.html ![]() You can download a free program that will chart curves based on your input. Vortex program
(from page 4) regards, T.Roc This post has been edited by TRoc on Sep 7 2007, 11:49 PM -------------------- I know Nothing. I looked all over to find it, but found it Nowhere. The funny thing is, it was right between 2 things, that I knew Everything about. It felt like forever, but really, it was no Time at all. |
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| jal |
Posted: Sep 8 2007, 01:57 AM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Hi Why Not?
I guess I'm trying to say too much at once. Put a colored dot/particle at the center of you CAD. The next dot/particle has got to be the diameter of one dot/particle away. So your first dot/particle will have 6 empty/white dots around it. (that is suppose to be the planck length distance.) Now, place the other five dot/particles on the outside of those empty dot. Each dot/particle must not come any closer than one dot distance from another dot/particle. Keep repeating the placing of dots/particles. See what pattern emerges. We should end up with a lot of things to discuss. Now, I'll go read those links. jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
| Why Not? |
Posted: Sep 8 2007, 05:07 AM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 327 Joined: 29-November 05 Positive Feedback: 53.33% Feedback Score: 1 |
Hey TRoc, Jal, and Wulf,
TRoc, thanks for the lead on number spirals and Vortex. The “problem” is that the number spirals require perfect squares on the “x” axis and assumes a square lattice. But I am going to read through the site in depth tomorrow and see if something triggers some insights in applying it to a hex lattice. Also, the concentric circles are not arbitrary at all. The radius of each concentric circle is an even integer multiple of the diameter of a unit circle plus one. So a concentric circle with a radius of 3 will define the radius of the first hexagonal packing, r = 5 for the second and so forth. Both the concentric circle and associated hexagon packing progressively increase as integer multiples of the unit circle diameter. I have drawn a similar picture with concentric circle radii of integer multiples without adding one, in which case the concentric circles intersect the center of each circle at the vertex of the hexagon defined by the packing defined by the same radius. In this case, at packing levels that are two times a prime of the form 6n+1 (14, 26, 38) there are an additional 12 unit circles within the hexagonal lattice that also have radii intersected by that concentric circle. In bith cases, the additional unit circles that are interannly tangential (or of the same radius) lay outside the hexagon associated with that radius. The only reason that I choose the first way was that I found it made comparisons of the area of the concentric circle to the additive area of the unit circles contained in the associated packing level easier. I am not sure why, but clicking on the pictures in the first post doesn’t seem to work to bring up an enlarged view anymore. If you right click and “Save Picture As” you will be able to zoom in and see the necessary detail. I did not draw in the hexagon sides... Jal, do you mean like this? Looks a bit like the picture in the link included in my last post... Mahalo This post has been edited by Why Not? on Sep 8 2007, 05:13 AM -------------------- Nothing is certain.
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| jal |
Posted: Sep 8 2007, 03:25 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Hi TRoc, Why Not? Why Not? Yes, that is the picture. With this picture you are representing a node/particle with the minimum possible distribution pattern that is allowed by applying the planck scale. You cannot bring anything closer to each other and still have any meaning. Do you agree? Remember that this is the 2d pattern that happened when we took a sphere and layed it out on a flat surface. Now the hard part... we got to bring 6 of those nodes/particles to a sphere. (remember that part about about the cube and 6j) We got to find a way of taking every one of those full and empty nodes/particles and arrange them into spheres. Those nodes/particles in the spheres cannot be any closer to each other than what we have just shown. (One planck length) A visual is worth a 1,000 words. I'm glad that you can manipulate CAD to help with this. The hard part might need a intermediate step for everyone to follow. You know how we represent a cube by drawing 6 squares on a flat surface. Can you superimpose those 6 squares over the nodes/particles? I'm not talking down to you .... I want to make sure that a 12 year old can follow this discussion. I'm sure that you are way ahead and realize that we will then go to packing 12 spheres. You will see that there will be a need to discuss options before that happens. ----------- TRoc!
That was just an awsome original way of demonstrating relations/mathematics. -------- jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
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| Why Not? |
Posted: Sep 9 2007, 03:17 AM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 327 Joined: 29-November 05 Positive Feedback: 53.33% Feedback Score: 1 |
Hey Jal,
Sort of... I am inclined to think that 2pi times 3 times the Planck length is the minimum possible propagation distance required to quantify transmission of information. Or, when bringing objects together, represents the point where it is impossible to distinguish 2 from 1. Since the number assigned to a “unit” is completely arbitrary, the unit circles in my original post can be considered to have a radius of 3Lp. As such, the concentric circles represent propagations through spacetime from the center outward. This post has been edited by Why Not? on Sep 9 2007, 03:18 AM -------------------- Nothing is certain.
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| DavidD |
Posted: Sep 9 2007, 03:47 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1881 Joined: 2-September 07 Positive Feedback: 26.56% Feedback Score: -151 |
What is behind planck constant? What is smallest elements of universe? Atoms? No. Electrons? No. Quarks? No. Why there is no gravity explanation? How gravity can be: between big objects, between small particles... But how to explain gravity? How to explain magnetism? This is also is a gravity... What we will se if we will be at scal of planck constant? Do we see then gravity? Waht we will see? Do we see, what is photons of what consist electrons, quarcks and over particles? At what speed interactions going? Why nothing smaller can't be than plank constant? And of what consist all matter in Universe? I have one teory, what all matter in universe consist of small bals, which size is a plank constant size. This balls is Ideal sphere shape. Nothink can't be smaller than this balls. And this balls is bits in Universe language encoding and decoding. This balls is like metallic balls, so don't has any gravitation or electromagnetic property. This balls in all Universe and Cosmos is in infinity amount. There is only one question: what is distance between those balls? Maybe at all no distance between those balls (those balls is infinity near each over)? From where goes energy to compute with those balls? Energy goes from infinity Nature of cosmos (space). Becouse cosmos is infinity in space, those balls can't take only one position and don't move, becouse over balls, from infinity cosmos push those balls and this process is infinity. So those balls never is in the same place, but always move from point to point. If we took one of those balls, if we looking at him, what he will do, and if we nothing do to him, we will see like overs balls pushing this ball in many directions. This one ball can't be deformed. All matter is result of those balls interaction between each over. Photons, quarks, protons and all over matter, particles is just result of computation, interaction of those balls. If you think, what Nature has gravition itself and what one particles consist of over particles, those over particles of over, which explain gravity, when don't you think, what this is a little bit absurd? Becouse then we can go to infinity small particles and never answer to question what is gravition and electromagnetism. This teory can give answer, what is gravitation and electromagnetism. It is interaction between balls, which in some way encode and decode information in what phorm in which we see all world. Like computer working with bits, can give to you such wide amount of information, in this sense working and universe, having only ball (1) and emptiness (0).
Opposite to this teory can be teory about liquid (fluid), which can be divided into infinity many parts. This post has been edited by DavidD on Sep 9 2007, 04:05 PM |
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| jal |
Posted: Sep 9 2007, 04:16 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1459 Joined: 23-October 05 Positive Feedback: 64.29% Feedback Score: -15 |
Hi Why Not? I want to make sure that we communicate properly and that a 12 year old can follow what we are saying. I want to clarify your point.
Since we are dealing, at this stage, with a 2d surface we can say circles. The circumference of a circle is 2pi therefore, when you said "2pi" did you mean a circle? Are you saying that "the minimum possible propagation distance" requires 3 circles? This would be 3 Planck lengths as represented by the diameters of the circles in the hex pattern. The second part of your statement, (the point where it is impossible to distinguish 2 from 1), is harder to understand or to "see" in the hex pattern.
Hummm-m-mm! That is moving too fast for me. If we go back to having the separating distance as one planck length and the diameter as one planck length then this agrees with the drawings. ------------------ We still got to bring this 2d information into a 3d configuration. Here is what I did with my inferior paint program. ![]() There is a problem when trying to go from a hex pattern in 2d to a cubic pattern in 3d. Here is my First try..... I don't know if it shows but I flatten a cube over the hex pattern that you drew. If I was to fold the cube then I would not be able to return to the original pattern of having a node/particle at the center of each face of the cube. Do you see that the two nodes/particles on the arms of the cross have been split in half? Second try .... That is what you see on the right side. I just selected a square with the node/particle at the center and glued them together. Looks pretty good heheh BUT it does not work either ... the problem has been moved from the node/particle to the space (planck distance). Did you try it? Do you see the problem? ------------ Hi DavidD I see you posted while I'm preparing my post so I will answer.
You should be able to "see" from the drawing that if you assume a minimum length then your statement is false. You should try to develop your theory in another thread. This thread is exploring the possibility of having a minimum length and at this stage the assumption is planck length. Nothing goes smaller than planck length in this thread. and we will see if there is an emerging structure. jal -------------------- Moved 10 June 2008
JAL'S BLOG http://www.physicsforums.com/blog.php?b=2 |
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| DavidD |
Posted: Sep 9 2007, 04:38 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1881 Joined: 2-September 07 Positive Feedback: 26.56% Feedback Score: -151 |
But what if this ball is plank lenght? And distance between those balls no exist. Those ball then still can move. And becouse cosmos is infinity and balls moving is analog, in principle one ball can move infinity small amount of lenght... Also I can add, that friction between those balls not exist, becouse balls is ideal sphere shape. And I don't see why those balls can't move faster than light. This post has been edited by DavidD on Sep 9 2007, 05:07 PM |
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