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Thread: Cartan spinor

  1. #1
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    Cartan spinor

    Élie Cartan (1869-1951) was one of the founders of modern theory of Lie groups, which have many applications in theoretical physics starting with Dirac equation for the electron of a relativistic quantum mechanics. In 1913, he discovered the most general mathematical form of spinors as linear representation of simple groups. These provided a linear representation of the groups of rotations in multidimensional spaces. Each spinor has 2 to the power of v components where the dimension n is 2v+1 or 2v. For n=4 of spacetime, every spinor have two components. Dirac used them to discover the existence of quantized spin for all elementary particles as well as for all elementary antiparticles. But theoretically, its lack of certain characteristics of affine geometry made them very difficult if not impossible to combine 4D spinors to 4D tensors of general relativity. All attempts to produce dimensional circle and orthogonal invariance have failed. Respectively, these violated Euclid’s 3rd and 4th postulates.

    The theory of spinors is based on the existence of isotropic vectors, zero length vectors such that the sum of squares of its components is identically zero. For the vector (a,b,c) in Euclidean 3-space (E³), it means that a²+b²+c²=0. And spinor (a,b) exists such that a=a²-b², b=i(a²+b²), and c=-2ab. These have two solutions for a and b in the complex domain. However, Cartan believed it is not possible to give a consistent choice of sign which will hold for all isotropic vectors in such a manner that the solution varies continuously with the vector. On the other hand, if a and b are real positive integers and a>b then the pair determine a unique Pythagorean triple or its reducible multiples. For example, a=2 and b=1 then a²-b²=3, 2ab=4, and a²+b²=5, and (3,4,5) is one unique Pythagorean triples widely used in land engineering design: 3²+4²=5². There are infinite number of Pythagorean triples and the question whether these contain all the prime numbers between 0 and infinity is worthy of further investigation.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Re: Cartan spinor

    Well here is a spinor for you Antonio:


  3. #3
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    Re: Cartan spinor

    That's a supernova spinor getting ready to explode or implode due to internal magnetic helicities entanglements. Thank goodness, it's only a virtual image inside this forum.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Smile Re: Cartan spinor

    Ying and Yang are your main man primary spinnors,all else arises from them.




    regards michael.
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    reveal herself?

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    Re: Cartan spinor

    If there is a will there is a way. Could that be the other way around?
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Smile Re: Cartan spinor

    Quote Originally Posted by AntonioLao View Post
    If there is a will there is a way. Could that be the other way around?
    Yes go in with Yang and out with Ying!




    regards michael.
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    Re: Cartan spinor

    go in with the yin and out with the yang would work just as well but together these double dualities become a quadruple entity of spacetime charges or double actions principle or square of energy.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Smile Re: Cartan spinor

    Quote Originally Posted by AntonioLao View Post
    go in with the yin and out with the yang would work just as well but together these double dualities become a quadruple entity of spacetime charges or double actions principle or square of energy.

    Absolutely,the square of energy would equal that of willed intent?




    regards michael.
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    reveal herself?

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    Re: Cartan spinor

    Speaking of yin/yang gentlemen.Linda just told me she wants to come back

  10. #10
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    Re: Cartan spinor

    double intentions or quadruple intentions? I going for an octuple intentions.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

 

 
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