Spacetime Topology Polysign

This thread's got drift alright.
Spoonfed wrote:
I looked at it a little more... You say "the special properties of + do
not extend beyond the two-signed domain." However, then you mention
that (*1)(anything) = anything. In this case, at least in
multiplcation, *1 takes the role of +1.
In the three-signed domain you are correct.
Also, in the three-signed case, you can use complex numbers to mimic
the properties.
"-1" = e^(i 2 Pi /3) = cos(120)+i sin(120)
"+1" = e^ (i 4 Pi / 3) = cos (240) + i sin (240)
"*1" = e^(i 6 Pi / 3) = 1 = cos (360) + i sin (360)

You probably already know this, but e^(i x), sin (x) and cos(x) as well
as less familiar functions... bessel functions, legendre polynomials
can be found with Taylor Expansion series.

e^x = x^0/0! + x^1/1! + x^2/2! + x^3/3! +...

It would be really neat if e^x meant something for P4 and up!

Plug an imaginary number in for x and you'll see a real and imaginary
part. The real part is the definition of cosine and the imaginary part
is the definition of sine.

Anyway, I just wondered if you were aware that there is something
similar in traditional mathematics to your multi-signed numbers in
traditional mathematics. Four signed numbers would have angles 90,
180, 270, and 360. Five signed would have angles 72, 144, 216, 288,
360. It came as some surprise to me when I realized that any root of
-1 could be described as a complex number.

You are imposing higher signs on a 2D space here. The polysigned
construction requires that four-signed numbers are 3D and five-signed
(P5) are 4D. Their angles come from the geometry of the simplex.
Projected down to 2D they could take on non harmonic values.
These systems do naturally generate unity roots and rotation.

But this isn't similar to yours for four-signed numbers, because in
your example you show four-signed numbers in three dimensions in a
tetrahedral pattern.
Right
Now it occurs to me that your three signed numbers and four-signed
numbers are linearly dependent. That is, one of your unit vectors can
be described as a linear combination of the other two. I don't know
whether that's important to you or not.

I don't see it this way. I really appreciate you spending energy on
this construction. I see P3 and P4 as two individual number systems
that are part of the same family. They are adjacent in the natural
progression that is the family of polysigned numbers. This is where the
product behavior goes haywire so it is of interest. The notion of a
linear relation between P3 and P4 is doubtful. I have spent some time
on this and P4 does look a lot like
P2 X P3 ( or R X C in traditional terms)
but I have yet to define a product on P3 X P2 that transforms exactly.
Still, If i do a 3D Mandelbrot plot of P4 it looks like a thick
Mandelbrot set as if it were true.
P2 comes out as the identity axis of P4 with P3 perpendicular to that.
But there is still some error. You can see this a bit in

http://bandtechnology.com/PolySigned/Deformation/DeformationUnitSphereP4.html
Sorry for the complicated and unfinished answer. How do you see linear
dependence in P3 and P4?

The discrepancies with the accepted construction are suggestive.
Traditionally a zero-dimensional quantity is considered to be a point.
That could use some clarification. Quantities have units, and that
entails dimension. A location might be zero-dimensional. Also, a
quantity might have zero uncertainty in its value, so in that way,
maybe could be considered zero dimensional.
That is still what one-signed numbers render out to, but there is a
little bit more there.
Can you add these one-signed numbers together? Or is the only
one-signed number zero?
Just enough to get time with it's arrow and without any graphical
measure.
*poof* what? how'd time get there? Because it is observed?

OK. This is an important one. I cover it pretty thoroughly at
http://bandtechnology.com/PolySigned/OneSigned.html
The polysigned numbers are intimately tied to dimensionality.
I take the one-signed numbers as the definition of zero dimensional.
We can do things like:
- 1.1 ( - 2.3 - 0.5 ) = - 3.08
in one-signed numbers.
Thinking in terms of superposition they can only get larger and larger.
So for example an integral of them should generate a nondecreasing
function.
Yet all the while when one graphs them they yield zero.
That is a funny thing about the cancellation law. You only actually
need it when you graph these things. All math can be done in any sign
level without ever invoking the cancellation.
In effect the components will keep building larger and larger values.
So long as they remain balanced you get a local answer.
The one-signed time correspondence is perfect because of this behavior.
It allows time to be part of spacetime without any graphical measure.
Do one-signed numbers have a magnitude? Yes:
| - 3.08 | = 3.08 .
But rendering that magnitude always yields zero.
To buy it you have to adopt the polysigned numbers and take them
literally.
One-signed numbers are time much much more convincingly than the real
numbers.
Also multidimensional spaces are developed without a Cartesian product.
These concepts are fundamental mathematics.
I am starting to believe that we exist not in a Cartesian product space
but in a particle product space.
Is it possible there might be more than one valid representation of the
same thing? This vaguely reminds me of the Lagrangian Method, which
greatly simplifies a lot of mechanics problems by replacing the
cartesian coordinate system with a system of generalized coordinates
based on the interaction of objects.
The polysigned numbers are nonorthogonal so there are inherently many
representations for the same geometric position. I'd like to coin a
phrase like 'minimally nonorthogonal' for they barely waste information
and it can even be argued that they reduce information over their
Cartesian counterpart. A 2D cartesian representation takes 2.2 chunks
of information whereas the polysigned equivalent can do it in 2.15
chunks where 1 chunk is a magnitude and 0.1 chunks is a bit that
becomes sign information. Sorry if that is cryptic. I can expound if
you wish.

This would allow for a polysigned
substrate whose product yields spacetime. This view may rely solely on
the superposition principle i.e. we need only look at particles two at
a time and allow that more are satisfied combinatorically.
In this context a sole distance exists between two particles, not four
independent distances.

Now you've lost me. Where do you get the idea anybody thinks there are
four independent distances between particles?

This is almost a rhetorical question. When we instantiate particle
positions in traditional spacetime representations we are forced to
assign four values. All of them are distances and they are independent
of each other hence the 4D spacetime model.
The traditional model is:
1D + 1D + 1D + 1D.
The polysigned model is:
0D + 1D + 2D.
To instantiate a particle position the tatrix requires six magnitudes.
Three of these can be disappeared by cancellation leaving three
magnitudes.
But there is another way and it gets around the nonisotropic quality of
the topology.
It uses relativity. Consider a universe with just two generic point
particles a and b. One solitary distance. This implies that:
| a1 | = | a2 | = | a3 | = | b1 | = | b2 | = | b3 |
where these are the 0D, 1D, and 2D components.
Each particle presents to the other particle in a relative fashion.
Claiming invariant distance on the topology leaves a few degrees of
freedom since the original format was nonorthogonal.
In 2D we see that there is a sign choice.
Let's say the distance between a and b is 1.234.
This means that in 2D the distance is 1.234 and in 3D it is 1.234.
a2 could be -1.234 or +1.234 or it could even be - 1.1 + 2.334, etc.
b2 is valued likewise. But ultimately this freedom is merely a choice
of two signs. It is a binary quality. In 3D the freedom is one of an
angle. Each particle can freely present these qualities to the other
particle.
This same procedure in the 1D + 1D + 1D + 1D would yield four binary
choices per particle. Not very sensible.
The polysigned approach generates 3.2 chunks of information and the
traditional approach generates 1.4 chunks. The distance is included as
one chunk. Very different.
Thinking this way can open up questions about d(a,b) versus d(b,a) but
this analysis assumes that they are equal.

Nicely this perspective generates binary charge
with axis out of generic point particles when taken relatively through
T3(P1 + P2 + P3). I can even argue spin 1/2 and spin 1 behavior is
exhibited in P4 when this invariance is invoked, though my claim is
shaky because I don't have enough understanding of spin.

To your original point...
If this product space is accurate it needs to be implemented in
arithmetical terms. Classical force equations look promising since they
provide a product relationship. However, the /r/r term is in conflict.

In conflict with what? The surface area of a sphere is proportional to
r^2. The part you are exposed to is 1 part of the surface area. As
you back off, the part you are exposed to decreases by 1/r^2
The polysigned numbers generate spacetime as a product relationship.
It is the product behavior that allows the claim.
So this transformation is a step in that direction.

Did you know that the force for an infinite cylinder is proportional to
1/r, and for an infinite plane is constant?
No I didn't know that. I don't follow the model. Are you instantiating
point charges on these structures? I don't see how that could yield a
constant force at any distance. What about a line? Does it blow up?

Consider the following transformation:
Y = 1 / ( 1 + X )
where X is traditional Cartesian distance and Y is now a measure
ranging from unity at what was the origin to zero at far away places.

Kind of like taking a train track that extends infinitely to the
horizon, and marking off the positions where you see it through a glass
window?

No. Oddly the tape measure looks almost the same. It's just that the
units read:
1/1, 1/2, 1/3, 1/4, ...
instead of:
0, 1, 2, 3, ...
Cut off the first inch and flip the numbers.


In effect we take a tape meaure, chop off the first inch, and flip it
upside down. It's the same basic tape measure slightly modified so that
now the force equation becomes:
F = q1 q2.

I'd have to ask what you started with. If F=(q1 q2)/x^2, this yields
(q1 q2 Y^2)/(1-Y)^2

Let's let r be in X so it is the usual value.
The classical charge force equation reads:
F = ( q1 / r ) ( q2 / r )
where q1 and q2 are charges that have a magnitude and a sign and r is a
distance.
We believe that charge is actually quantized so we could forgo some
more constants and treat q1 and q2 as just signs. i.e. each of them is
+ 1 or - 1. Let's do that.
But the polysigned numbers want a geometrical product relationship.
We can make this product a geometrical product by transforming distance
by
1 / X
but even neater is to go to
1 / ( X + 1 )
So in Y we get:
F = (q1( r + 1 )) (q2( r + 1 )).
For large r (which still has the sense of X) the 1 will become
irrelevant.
Since the distance is unsigned we can combine it into the q which is
currently just a sign to get
F = q1 q2
where q's magnitude is its distance in Y and q's sign is its charge.
We now have a geometric product that is equivalent to the classical
equation for large r.
If we are discussing point particles there should not be any conflict.
Infinity is gone at near distances and at far distances.
I have no idea what this would say about strong and weak forces.
I'm just trying to get polysigned physics working.
I also like the transform because it has an intuitive sense about it.
Unity is local and zero is meaningless.

-Tim

.

Relevant Pages

  • Re: Spacetime Topology Polysign
    ... maybe could be considered zero dimensional. ... but in a particle product space. ... A 2D cartesian representation takes 2.2 chunks ... In this context a sole distance exists between two particles, ...
    (sci.physics.relativity)
  • Re: Stephen Hawking on waves and particles
    ... on Sun, ... Do you deny the distance from A to A is zero? ... The distance from A to A is indeed zero in any reasonable metric. ... given that we have no reason to believe a particle, ...
    (sci.physics)
  • Re: A puzzle
    ... > (distance, velocity, time), they are linked in that one must be the ... > where any particle had traveled or where it was headed? ... > about the path of a particle as it travels through space. ...
    (sci.physics.relativity)
  • Re: Does the Electromagnetic field have a Gravitational field?
    ... >> Wheeler particle interaction, without the traditional ... > at a distance theory of gravity. ... that showed no need for gravitational energy sourced ... a test particle, ...
    (sci.physics.research)
  • Re: Effectuationism and Value
    ... > reference giving the actual equation in terms of r'; the distance taking ... > retarded time into account. ... > The force will depend on the position the particle was at that time if ... Since I used a standard term I naturally assumed ...
    (sci.physics.relativity)
Quantcast