The red is the uniform uncertainty is a measurement coming form an integer deviation counter of rate cosh.
I we think in terms of three cosh curves, rotated about each of he three axis. We count N from 1 to N along the axis specified by 1,1,1 to x,y,z. t each steps the deviation couns increment by rate, and counted rate steps around its axis of the conic and colors it.
As the axis is marked from 111 to xyz, all the deviation counters will spiral up that axis marked a discrete ribbon about their rotated cosh.
The Markov node of the even odd variety must be light modes. Boson statistics can recolor the surface, from odd then back to even, so light is a deviation bundle counting N, the light capacity of free space is an N density.. Light moves because it splits into a positive curvature followed by a negative curvature, the magnetic rotation of the color operator opposite of each other. It moves because it removes deviations in N.
Charge isn't negative? No curvature is is reversed on the operators. The magnetic field on the proton, is accurate, and reversed and 'stitches' up the points along the rotation where deviation counts would otherwise be maximum.
So the vacuum can curve, and is favorable to the natural curve of compression. The vacuum element can spin in either even/odd, unless we are talking quarks. Charge is rae of curvature. The vacuum is absolute positive, no negative, no zero. Most of the grunt work done by the magnetic. When N is assume infinite, or always locally sufficient, then light appears as a closed surface moving across some N of minimum divergence.
Quarks are a faild attempt to jump into the 5,y,z branch of Markov. Not enough N. It is under sampled, does not close and quarks are escape partitions. Pauli exclusion is true because the quarks will not close. So there is constant flipping between boson and fermion. Light cannot keep up with spin and spin destabilizing will spin segment with opposite spin, the N partition lets light in the first catch up and spin partition becomes charge partition. Too few N for two operators, two many for one. The under sampling comes to 3/2, it should be 2,
To close the surface you need N, otherwise you keep recoloring or keep losing partitions in N.
I we think in terms of three cosh curves, rotated about each of he three axis. We count N from 1 to N along the axis specified by 1,1,1 to x,y,z. t each steps the deviation couns increment by rate, and counted rate steps around its axis of the conic and colors it.
As the axis is marked from 111 to xyz, all the deviation counters will spiral up that axis marked a discrete ribbon about their rotated cosh.
The Markov node of the even odd variety must be light modes. Boson statistics can recolor the surface, from odd then back to even, so light is a deviation bundle counting N, the light capacity of free space is an N density.. Light moves because it splits into a positive curvature followed by a negative curvature, the magnetic rotation of the color operator opposite of each other. It moves because it removes deviations in N.
Charge isn't negative? No curvature is is reversed on the operators. The magnetic field on the proton, is accurate, and reversed and 'stitches' up the points along the rotation where deviation counts would otherwise be maximum.
So the vacuum can curve, and is favorable to the natural curve of compression. The vacuum element can spin in either even/odd, unless we are talking quarks. Charge is rae of curvature. The vacuum is absolute positive, no negative, no zero. Most of the grunt work done by the magnetic. When N is assume infinite, or always locally sufficient, then light appears as a closed surface moving across some N of minimum divergence.
Quarks are a faild attempt to jump into the 5,y,z branch of Markov. Not enough N. It is under sampled, does not close and quarks are escape partitions. Pauli exclusion is true because the quarks will not close. So there is constant flipping between boson and fermion. Light cannot keep up with spin and spin destabilizing will spin segment with opposite spin, the N partition lets light in the first catch up and spin partition becomes charge partition. Too few N for two operators, two many for one. The under sampling comes to 3/2, it should be 2,
To close the surface you need N, otherwise you keep recoloring or keep losing partitions in N.
No comments:
Post a Comment