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Sunday, March 10, 2019

The quantum shortest distance

This drawing (which happens to be my registered trademark) shows that the shortest distance between two quantum states is a right triangle.
It also shows that to get the recognized symetries the positive and engative values must be ofset and the amount of the offset is goverened by the percision with which the F-series corresponds to the circle or in the words of AuT theory, the extent towhich the denominator of pi yields the fibonoacci series which are show related by this drawing intersecting the two with relative percision.
In this case, you have f(n) for n, n+1, n+2 and n+3.  One more circle out would represent +4 which is the black hole state which is not required for the analysis.
If you draw a straight line from the center to the point of intersection, you divide the f series curves into right angles.
This has to be used to solve bells inequality in terms of AuT.

Today is a swimming day in concept.  Since the pool doesn't open till 1 it is hard to predict the outcome there.  Ahh, but what an icing on the cake type of thing if I get a good workout in!
The offsets in this are pretty cool, because they perfectly align the vitruvian man with the AuT man.
That's what AuT is.  It takes stuff like Einstein, Newton, Davinci, Parmenides, Schrodinger, Bell, and Fibonacci adn ties them all up together with a nice little bow.  Its like bondage-physics thing.

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