Invariant intervals

The spacetime interval is invariant over the Lorentz transformation (LT). The following is a proof of this for the inverse LT with length space 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):

The invariant interval is

Expand the squares and cancel the middle terms to get:

The timespace interval is invariant over the dual Lorentz transformation (DLT). The following is a proof of this for the inverse DLT that follows with time axes x, y, and, z; stance axis r (placeline), lenticity u, and maximum lenticity κ, along with ρ = ℓ/k and λ = 1/√(1 − ρ²):

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

The invariant interval is

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

$=\lambda ^2(\Delta x'+ \rho k \Delta r')^2 +\Delta y'^2 +\Delta z'^2 -\lambda ^2(k \Delta r'+\rho\Delta x')^2$

$=\lambda ^2(\Delta x'^2 (1-\rho^2) -k^2\Delta r'^2(1-\rho^2)) +\Delta y'^2 +\Delta z'^2$ Expand the squares and cancel the middle terms to get:

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

$=\Delta x'^2 +\Delta y'^2 +\Delta z'^2 -k^2\Delta r'^2$

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

$(\Delta s)^2 = (c\Delta t)^2 -(\Delta r)^2 \; \textup{and}\; (\Delta \emph{l})^2 = (\Delta t)^2 -(k \Delta r)^2$

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

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.

If c = κ = 1 (without regard to units), then these invariant intervals are nominally the same:

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