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Doubled of Minkowski space and spinor wave function

  1. Apr 29, 2014 #1
    First of all note that 8-dinensional Finsler space [itex](t,x,y,z,t^*,x^*,y^*,z^*)[/itex] preserving the metric form
    \begin{equation}
    S^2 = tt^*-xx^*-yy^*-zz^*,
    \end{equation}
    actually presents doubled of the Minkowski space.

    Then the solution with one-dimensional feature localized on the world line of doubled of Minkowski space [itex](x, y, z, t,t^*,x^*,y^*,z^*)[/itex], which in the distance from it tends to the vacuum potential, should be considered as particle-like solutions. Moreover, if we are interested in particle-like solutions in which the static part of the line potential features has symmetry of the equator of seven-sphere, and the dynamic characteristics of the line is wound on this equator, then the space which is orthogonal to this line (or rather - a congruence of lines) features can be represented by spinor wave function:
    \begin{equation}
    \begin{cases}
    \psi_1= \frac{z_1}{|z|}e^{iS},\\
    \psi_2= \frac{z_2}{|z|}e^{iS},\\
    \psi_3= \frac{z_3}{|z|}e^{iS},\\
    \psi_4= \frac{z_4}{|z|}e^{iS},
    \end{cases}
    \end{equation}
    where [itex](z_j = x_{2j- 1} + ix_{2j})_4[/itex] --- a point north pole of the sphere with radius [itex]|z| = \sqrt{z_1\bar{z}_1 + \cdots + z_4\bar{z}_4}[/itex], and [itex]S = k_{x}x + k_{y}y + k_{z}z-k_{t}t[/itex] --- this is the path length features in the space [itex](t, x, y , z, t^*, x^*, y^*, z^*) [/itex]. Note also that if we are interested in particle-like solutions with the symmetry of the pseudo-sphere radius [itex]\sqrt{z_1\bar{z}_1 + z_2\bar{z}_2 - z_3\bar{z}_3-z_4\bar{z}_4}[/itex], but their internal symmetry is broken by the inequality [itex]\sqrt{z_1\bar{z}_1 + z_2\bar{z}_2} \gg\sqrt{z_3\bar{z}_3 + z_4\bar{z}_4}[/itex], we can limit the two-component spinor describing the symmetry on three-dimensional sphere with radius [itex]\sqrt{z_1\bar{z}_1 + z_2\bar{z}_2}[/itex].

    Whether it is worth discussing?
     
  2. jcsd
  3. Apr 29, 2014 #2

    micromass

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    Please stop using physicsforums to try and discuss this.
     
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