X-Message-Number: 20697
From: 
Date: Mon, 23 Dec 2002 16:35:25 EST
Subject: Non-local force equation: How it looks

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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.

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