iSoul Time has three dimensions

Invariant intervals

The spacetime interval is invariant over the Lorentz transformation (LT). The following is a proof of this for the inverse LT with spatial axes x, y, and, z; temporal axis t (time line), velocity v, and maximum velocity c, along with β = v/c and γ = 1/√(1 − β²):

\Delta x=\gamma (\Delta x'+\beta c\Delta t');\; \Delta y=\Delta y'\; \Delta z=\Delta z';\; c\Delta t=\gamma (c\Delta t'+\beta \Delta x').

The invariant interval is

(\Delta s)^2 = (c\Delta t)^2 -(\Delta x)^2 -(\Delta y)^2 -(\Delta z)^2

=\gamma ^2(c\Delta t'+\beta\Delta x')^2 -\gamma ^2(\Delta x'+ \beta c\Delta t')^2 -\Delta y'^2 -\Delta z'^2

Expand the squares and cancel the middle terms to get:

=\gamma ^2(c^2 \Delta t'^2(1-\beta ^2) -\Delta x'^2(1-\beta ^2)) -\Delta y'^2 -\Delta z'^2

=\gamma ^2(c^2 \Delta t'^2(1/\gamma ^2) -\Delta x'^2(1/\gamma ^2)) -\Delta y'^2 -\Delta z'^2

=(c\Delta t')^2 -(\Delta x')^2 -(\Delta y')^2 -(\Delta z')^2.


The spacetime interval is invariant over the dual Lorentz transformation (DLT). The following is a proof of this for the inverse DLT follows with temporal axes x, y, and, z; spatial axis r (stance line), legerity u, and maximum legerity κ, along with ζ = u/κ and λ = 1/√(1 − ζ²):

\Delta x=\lambda (\Delta x'+ \zeta \kappa \Delta r');\; \Delta y=\Delta y'\; \Delta z=\Delta z';\; \kappa \Delta r=\lambda (\kappa \Delta r'+\zeta \Delta x').

The invariant interval is

(\Delta w)^2 = (\Delta x)^2 +(\Delta y)^2 +(\Delta z)^2 -(\kappa \Delta r)^2

=\lambda ^2(\Delta x'+ \zeta \kappa \Delta r')^2 +\Delta y'^2 +\Delta z'^2 -\lambda ^2(\kappa \Delta r'+\zeta\Delta x')^2

Expand the squares and cancel the middle terms to get:

=\lambda ^2(\Delta x'^2 (1-\zeta ^2) -\kappa ^2\Delta r'^2(1-\zeta ^2)) +\Delta y'^2 +\Delta z'^2

=\lambda ^2(\Delta x'^2 (1/\lambda ^2) -\kappa ^2\Delta r'^2(1/\lambda ^2)) +\Delta y'^2 +\Delta z'^2

=\Delta x'^2 +\Delta y'^2 +\Delta z'^2 -\kappa ^2\Delta r'^2.


Let Σ (Δri)² = (Δr)² and Σ (Δti)² = (Δt)². Then the two invariant intervals are

(\Delta s)^2 = (c\Delta t)^2 -(\Delta r)^2 \; \textup{and}\; (\Delta w)^2 = (\Delta t)^2 -(\kappa \Delta r)^2

These two invariant intervals are proportional if κ = 1/c:

(\Delta w)^2 =(\frac{\Delta s}{c})^2 = (\Delta t)^2 -(\frac{\Delta r}{c})^2 = (\Delta \tau)^2

In that case, the invariant interval for proper time is the same for both 3+1 and 1+3 dimensions. However, 3D proper time is limited to space-like intervals, while 3D proper distance is limited to time-like intervals. Since space-like intervals require superluminal speeds, they are considered impossible.

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