X-Message-Number: 20697 From: Date: Mon, 23 Dec 2002 16:35:25 EST Subject: Non-local force equation: How it looks --part1_d2.218346ab.2b38db9d_boundary Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Mostly for archival purpose: nonlocal force equation Recall of the preceeding episodes: Quantum mechanics is built on a space of points, when we try to expand it into the relativistic domain with time as a dimension, we fall on the spinor formalism. These are described by clifford algebras dealing with non local spaces. The elementary object is no more a point, it is an extended domain, each such domain contains and infinite number of points and we have a towering number of quantum theories: Dirac, Klein-Gordon, Dual, SUSY, string, brane, p-brane, M-theory,... Even an infinite set of such theories would not exhaust the possibilities as that infinity would be countable when the number of point in an extended domain is not. The proposed solution was then to take on one side the clifford domain and remove from it a finite number of points forming the basic element of a classical quantum space. From here, I have assumed that there is a translation formula between these domains. Something in the clifford space would exercise an influence on the quantum domain, that is we would see a quantum force with origin in the clifford space. That force would be described by a covariant derivative similar to the ones produced by geometrical curvature or group symetries. Because things would work the two ways, a quantum force from group symetry such the electromagnetism U(1) would produce a clifford force described by a covariant derivative too (geometrycal curvature -gravitation- would too produce such a force); So, the equation is: Covariant derivative on clifford space = covariant derivative on quantum space. A covariant derivative is an ordinary derivative *and* a quadratic term called gamma. In the simplest case, that of geometrical curvature, gamma is the product of an order one derivative by an order two one. So gamma is a three indice object: 1 for the first derivative and 2 for the second. Because second order derivatives are symetrical, the 2nd and 3rd gamma indice can be exchanged, they are symetrical. That is not true for more general gamma don't coming from curvature. There we can hope no pity from the formula: The right hand side is "compressed" ( a domain crammed into a point) so there will be no simple symetries. Assume we want the formula for non-relativistic quantum mechanics, the so-called first quantification. Only one point will be removed from the clifford domain. Gamma will have only 3 dimensions and so 3x3x3 = 27 coordinates. There is a special problem:How does we know that we are in quantum space? Why the missing part of clifford must be quantum? why not geometrical curvature for example? The truth is that a translation formula must define both, the starting and destination space. The quantum space is defined by a metric, a way to mesure distances built on the sum over all variable value of the product of a function and its complex conjugate (the same with inversed imaginary part). A function is a zero-form, the dual of a point. If gamma goes to a point, then a quantum space needs two: The quantum space is built from two superimposed points... That make gamma unlike what is seen in covariant derivative limited to a single space: Here, gamma is not a simple name, it is a vector index. Strange vector, defined by two points at the same place. So, gamma is a null vector, at least in quantum space. In fact this mathematical object is a rank 4 operator with one vector index and three differential ones. If the used space basis is altered, one index will trasform with a law and the others with a different law. It is rather uncommon to find two different tranformation laws, depending on which index we look at in a single operator. What is on the left side of the equation? To put things in the simplest possible frame, I assume the non local space is made of only two points: one field source and a sink. To be sure, there is no way to move one point to the other, so we could as well speak of two equivalent points, one on each side of a separator. Seen from far away, these points may be fused into one, so there is an ordinary covariant derivative. Assume we label the points l for left and r for right. If we look at them from a short distance they can be moved independently and a derivative may be applied to them, assume it is dl and dr. On the left hand side of the equation we must so have a sum of four elements: dl + dr + the ordinary derivative + gamma. The ordinary derivative is a single index quantity when there is no l-r problem. Here, we must start to multiply dl by dr and then derive the original index by l and then r. This is a five indice quantity with six way to contract it into the single index on the right hand side of the equation. Only four quantities are independent, so the null vector there is in four dimensions or only two if l and r are symetric. Gamma is longuer, it is the product of dl, dr, the drdl derivative of the order one derivative and the drdl derivative for each indice of the order two derivative with a mixed drdl derivative. There are ten independent ways to contract that object into an ordinary gamma, five if there is a symetry between l and r. The space of the gamma vector is so 10 dimensional. In index notation the ordinary derivative would be: Dabcde and gamma: Gabcdefghijklm, the total derivative would be: l + r + Dabcde + Gabcdefghijklm. D being a 4 dimensional vector index and G a 10 dimensional one. Counting everything, there are 22 indice to play with in that derivative. What are these indice is another matter. Yvan Bozzonetti. --part1_d2.218346ab.2b38db9d_boundary Content-Type: text/html; charset="US-ASCII" [ AUTOMATICALLY SKIPPING HTML ENCODING! ] Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=20697