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# Invariant intervals

The “spacetime” (length-time) 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):

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

The invariant interval is

$(\Delta&space;s)^2&space;=&space;(c\Delta&space;t)^2&space;-(\Delta&space;x)^2&space;-(\Delta&space;y)^2&space;-(\Delta&space;z)^2$

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

Expand the squares and cancel the middle terms to get:

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

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

$=(c\Delta&space;t')^2&space;-(\Delta&space;x')^2&space;-(\Delta&space;y')^2&space;-(\Delta&space;z')^2.$

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

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

The invariant interval is

$(\Delta&space;w)^2&space;=&space;(\Delta&space;x)^2&space;+(\Delta&space;y)^2&space;+(\Delta&space;z)^2&space;-(\kappa&space;\Delta&space;r)^2$

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

Expand the squares and cancel the middle terms to get:

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

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

$=\Delta&space;x'^2&space;+\Delta&space;y'^2&space;+\Delta&space;z'^2&space;-\kappa&space;^2\Delta&space;r'^2.$

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

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

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

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

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.

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

$(\Delta&space;s)^2&space;=&space;(\Delta&space;w)^2&space;=&space;(\Delta&space;t)^2&space;-(\Delta&space;r)^2.$