Repost from SCI.p.Research−Matrix Determinant as Path Integral?
Repost from SCI.p.Research−Matrix Determinant
as Path Integral?
Source: http://sci.tech−archive.net/Archive/sci.physics/2007−08/msg00265.html
• From: Douglas Eagleson
• Date: Fri, 03 Aug 2007 05:29:37 −0700
I will top or front comment. The original post follows.
It is difficult to convince people as it is a modern technique of
matrice. ALL matrix as a third abstracted relation of set element to
location is termed matrice. So inclusive of all graph theory was the
matrice. A study of the entire works of Hamilton is a good
introduction to this topic.
Hamilton literaly used quaterions to talk about his quest.
A matrix has a number locations associated with the set element. That
is the property unique to a matrix. It make it different from a
number without location relationship.
I can likely answer the question of correct usage of matrice, but open
the floor to debate. A method was suggested in the original post.
And the rules for speculation prevented the moderators of ..research
from obtaining an answer.
Here are the basic benefits of the method.
1. Infinite renormalized.
A single benefit appears. A means with renormal as operated element
appears. A selection of element size as opposed to sovled equation,
causes the solution. This is a major, major benefit. So I will
speculate about this topic to begin with.
An integral as opposed to the path integral will be used now. All
integral have solved elements called dx. This is simply a usage of
the basic formula. Normally dx is caused an infinite limit and the
function solved. In matrice all dx are given a constant, is it the
average dx for the integral top and bottom limits of integration, or a
true differential constant of third relative change over the span?
Sorry it is the second usage. A perfect integral is obtainable in
matrice.
Repost from SCI.p.Research−Matrix Determinant as Path Integral? 1
Repost from SCI.p.Research−Matrix Determinant as Path Integral?
This is the goal, I claim it obtained.
y=f(x)
For any function define the matrice as integral.
answer= determinante(cross product)matrix
matrix= |dx1 area|
|dx2 area|
|dx3 area|
|dx4 area|
This is Hamilton's matrix usage. Hamilton in my thinking discovered it
in western lands. A cross product of the matrix determinante with the
original matrix causes!!!!!!!!!!!!!!!!!!!!!!!
It defines his matrice. Without discussing this great discovery here
is the means of integral answering.
Matrix eigen as answer. AN Eigen is the term for the left hand side.
It is not an eigenvalue, but is derived from one.
SELECT Eigen.
And with Hamilton's matrice the selection ALWAYS appears a sufficincey
of dx limits!
For y=f(x) place the eigenvalues equal to the functions y at the
dx1,2,3,4 integral limits!.
A selected eigenvalue causes element area to equate. A literal search
will obtain a correct determinate, and thereffor integral.
Sufficiency is not the size of the Eigen, it is the limit selected.
So automatic limit number appearance is built in.
That is the usage. Beleive it or not.
The original poster wonders of determinate theory, and this answer
distills the wonder.
Lets get this exactly clear:
And eigenvalue set will cause four function values, each an x value of
Repost from SCI.p.Research−Matrix Determinant as Path Integral? 2
Repost from SCI.p.Research−Matrix Determinant as Path Integral?
the four dx elements.
When an independent four set matchs, a visual match, happens the Eigen
is the integral.
Independent means to have each x be the x−axis value for the given
function eqi−division of limits of integration.
Eigen as a term is a new term dervied from eigenvalue and is
technically a third order of logic use of matrix location to solve.
*******************Original post*******
This post is a little dreamy and may not make sense, so please bear
with
it. On the other hand, there might be something to it.
The other day I noticed that the determinant of a matrix looks like a
sum−over−histories. You can think of it as the sum of all the paths
from
the top to the bottom, and if you use complex numbers then you even
have
interference, but with the problem that you have the alternating {+1
and −1}
in the products. There is also a thing called the Permanent [see
wikipedia]
that doesn't have alternating signs, it just uses +1 the entire time.
You
can also define all sorts of things with other numbers on the unit
circle
like {1,i,−1,−i}, etc.
At first I thought it was interesting but not−so−great, until i
remembered
that the determinant is a group homomorphism. That means the kernel
is
mapped to the identity element of the image group. Now I often call
the
identity element the "do nothing" element, because it does just that:
nothing. For instance, 2*1 = 2, or 2+0=2, etc. Now the shortest
"distance"
between 2 points is a geodesic, and if its light, then its a "null
geodesic": it does nothing, and everything can be defined in reference
to
it. So fermats principle of least time, the concept of the geodesic,
and of
the fundamental frequency, and of quickest path to a win in a
stochastic
game, and least action all seem to be describable by the identity
Repost from SCI.p.Research−Matrix Determinant as Path Integral? 3
Repost from SCI.p.Research−Matrix Determinant as Path Integral?
element.
For instance, you can think of the fourier transform of the number 1:
You
have all negative and positive frequencies and they cancel out to a
dirac
delta function. Its the identity element of an additive group. You
wouldnt
get much energy coming over the radio at that frequency.
But the path integral cancels out to those same things. Perhaps the
determinant is some kind of path integral that is related to fermions,
the
permanent is one for bosons, and the other similar structures are for
anyons. I found some references on the internet, but they seemed sort
of
lofty and mysterious. Here are 3 questions that are possible a little
more
practical: if you have a system with a continuous spectrum, then
1. The matrix has a dimension
that is
continuous, so can the determinant be defined?
If it converged, it would be
an
integral of infinite products.
2. If it could be defined, would
the
resulting continuous−dimensional structure still be a group? [set of
matrices
with certain types of
determinant
= group.]
3. Do you need lebesgue theory
or just
the more familiar riemann integrals?
If you are an expert on this stuff, please explain.
JHS
.
Repost from SCI.p.Research−Matrix Determinant as Path Integral? 4