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Orthogonal Infinitesimal Forces

  1. Jun 19, 2004 #1
    If each spacetime point [itex] p_i[/itex] can be associated with a contant force [itex]f_i[/itex] then the interaction [itex] \sum_{i=1}^\infty f_i [/itex] between points can be described with the use of orthogonal forces.
     
    Last edited: Jun 19, 2004
  2. jcsd
  3. Jun 20, 2004 #2
    Orthogonality of forces is a consequence of when the metric of spacetime is exactly zero. When the metric is zero, each point can only interact with six forces and these forces formed two sets of three orthogonal forces.

    When the metric is exactly zero, spacetime is said to be continuous. And each point can only interact with orthogonal forces and with collinear forces.

    When the metric is the Planck length, then spacetime can be locally quantized by its discrete twist cyclic motion and satisfying the eight directional invariance's properties.
     
  4. Jun 21, 2004 #3
    Collinearity also allows a spacetime point to interact with only six points. But because of perspective, each spacetime point can "see" all the infinite minus six points of the whole fabric of spacetime. And to interact with any of these point, for only coplanar points, distances of the form [itex] \sqrt{ n^2 + 1}[/itex] for n=1 and as n approaches infinity must exist.

    The metrical value of the geodesic is assumed the value of 1.
     
    Last edited: Jun 21, 2004
  5. Jun 21, 2004 #4
    The general formula for distances between spacetime point is given by

    [tex] d_{mnl} = \sqrt{m^2 + n^2 + l^2} [/tex]

    where [itex] m [/itex] and [itex] n [/itex] and [itex] l [/itex] are the indices for the orthogonal vertical and horizontal planes found in one quadrant and the orthogonal planes with indices of (0,0,0) is defined as passing through the point itself at (0,0,0).
     
  6. Jun 22, 2004 #5
    When the shortest distance (geodesic) between orthogonal planes is given the value of 1 (e.g. 1 Planck length) then nearly orthogonal forces can be assumed to exist between spacetime points. But when the geodesic is exactly zero then the scalar products of orthogonal forces are zeros even though the forces have a constant value.
     
  7. Jun 22, 2004 #6
    The constancy of these orthogonal forces forms constant accelerations at the infinitesimal domain of spacetime. These accelerations vanish when the metric is exactly zero. And to make the metric zero, a quantity of [itex]c^2 t^2 [/itex] must be subtracted from [itex] d_{mnl} [/itex]. The new expression is equivalent to the spacetime interval of Einstein's theory of special relativity.

    [tex] d_{mnl} = \sqrt{ m^2 + n^2 + l^2 - c^2 t^2} [/tex]

    In a separate formulation, [itex] c^2 = \vec{a} \cdot \vec{r}[/itex] and [itex] a_m r_m + a_n r_n + a_l r_l [/itex] is not identically zero. But two of the indices must be zeros, (1,0,0), (0,1,0), (0,0,1). For the case where n=0 and l=0, then

    [tex] d_{m00} = \sqrt{ m^2 - m a_m t^2} [/tex]
     
  8. Jun 22, 2004 #7
    Since the indices [itex]m, n, l[/itex] take on only integer values, the distances between spacetime points are said to be quantized.
     
  9. Jun 23, 2004 #8
    To complete the structure of spacetime, the following equation is needed:

    [tex] d_{0n0} = \sqrt{n^2 - n a_n t^2}[/tex]

    These two distances formed a Hopf ring although individually, it's more like part of a hyperbola.

    A 3rd form [itex] d_{00l} = \sqrt{l^2 - l a_l t^2}[/itex] exists and random selections by nature's choice which two forms is taken out of the three possibilities led to two distinct topologies of the dynamic Hopf ring.
     
  10. Jun 23, 2004 #9
    Because there can only be two distinct metrics ([itex] \psi_E [/itex] and [itex]\psi_B [/itex]) of spacetime, there should be only two fundamental orthogonal forces ([itex] \phi_E [/itex] and [itex] \phi_B [/itex]). In a minimum-maximum configuration, these are equivalent to the electric force and the magnetic force of the vacuum. The existence of magnetic permeability ([itex] \mu_0[/itex]) and electric permittivity ([itex]\epsilon_0 [/itex]) of the vacuum is a proof that these forces do exist.

    And the vector-scalar products of these metrics and forces give the square of energy, another quantum of nature.

    [tex] E^2 = \psi_E \times \phi_E \cdot \psi_B \times \phi_B [/tex]
     
  11. Jun 24, 2004 #10
    The assumption that each spacetime point carries with it a constant magnitude force raises the question about the existence of triangular forces (force configuration bounded by three spacetime points). The lattice structure of spacetime at the fundamental level indicates that the minumum bounding points is four. If there are only three points as the vertices of an equilateral triangle, the distortion of the spacetime structure becomes tremendous which will give rise to very high curvature ripping the fabric of spacetime.
     
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