Belt Networks

A network of belts would be equivalent to a network of directed graphs.

A half twist in a belt corresponds to a directed arrow.

In particular a sub structure rotating 360 degrees would reverse the arrows going into the substructure:
 

Can we be more clear on graph equivalences?


Therefor such a patch as fermionic structure. But what are fermions? (Perhaps all nodes are fermions.)


Comments

Post a Comment

Popular posts from this blog

Vortex Solution to Navier Stokes Equations

Image
This is a solution I found to the Navier Stokes equations . It uses the flow lines of the Hopf Fibration (which is basically just a way of covering the entire 3D space in circles). The flow lines can be seen in this image below: You can see my paper here. Solution to Navier-Stokes equation . Unfortunately I won't get a million pounds for winning the Clay Millennium prize because this is a compressible not incompressible solution. :( Oh, well. Nice try! I think it's pretty nice anyway. :) Also it uses the nice identity: which paramaterizes the Pythagorean quadruples!
Read more

Distance Geometry in Superspace

We can define a distance invariant under superspace transformations: \[ R^2 = |x-y|^2 + \theta \gamma_\mu(x-y)^\mu \zeta \] For D space-time dimensions and E Grassman dimensions, does this satisfy a Cayley-Menger type equation? In 2 dimensions we know the volume of a tetrahedron is zero. Or: \[ \det \begin{bmatrix} 0 & R_{12} & R_{13} & R_{14} & 1\\ R_{12} & 0 & R_{23} & R_{24} & 1\\ R_{13} & R_{23} & 0 & R_{34} &1\\ R_{14} & R_{24} & R_{34} & 0 &1 \\ 1&1&1&1&0 \end{bmatrix} =0 \] What about when D=11 and E=32 ? Simplest is D=1, E=2 with (?): \[ R^2 = (x-y)^2 + (x-y)(\theta \zeta + \zeta^* \theta^* ) \] with for example: \[ R^4 = (x-y)^4 + 2 (x-y)^3(\theta \zeta + \zeta^* \theta^* ) + 2(x-y)^2( \theta \theta^* \zeta \zeta^* ) \] \[ R^6 = (x-y)^6 + 3 (x-y)^5(\theta \zeta + \zeta^* \theta^* ) + 6(x-y)^4( \theta \theta^* \zeta \zeta^* ) \] Then the area of the triangle is 0 (or an
Read more