Lorentz-like transformation for time-distance domain

Lorentz-like transformations for time-distance domain

What follows are derivations of the Lorentz transformation for the time domain and a Lorentz-like transformation for the distance domain.

Time Domain

Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with velocity v with respect to O in the positive x-direction. The origins of O and O′ initially coincide with each other. A light signal is emitted from the common origin and travels as a spherical wave front. Consider a point P on a spherical wavefront at a length rx and rx from the origins of O and O′ respectively.

For round-trip motion, the mean round-trip speed of light, c, is the same in all inertial frames, so for the point P with length vector r = (rx,0, 0) and

rx = ct, and rx′ = ct′.

Consider the standard Galilean transformation with a factor γ, which is to be determined and may depend on β, where β = v/c:

rx = γ(rx − vt) = γ(rx − βct) = γ(rx − βct) = γrx(1 − β).

The inverse transformation is the same except that the sign of β is reversed:

rx = γ(rx′ + vt′) = γ(rx′ + βct′)= γ(rx′ + βct′)  = γrx(1 + β).

Multiply these two equations together to get

rxrx = γ² rxrx(1 − β²).

Divide out rxrx to get

γ² = 1/(1 − β²),

or γ = 1/√(1 − β²).

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Consider what happens if this approach is applied to the independent time t times the constant c.

ct′ = rx = γ(ct − βrx) = γ(rx − βrx) = γrx(1 − β).

The inverse transformation is the same except that the sign of β is reversed:

ct = rx = γ(ct′ + βrx) = γ(rx′ + βrx) = γrx(1 + β).

These are exactly the same equations as before, so the result will be the same.

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Distance Domain

Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with rapidity p with respect to O in the positive x-direction. The origins of O and O′ initially coincide with each other. A light signal is emitted from the common origin and travels as a spherical wave front. Consider a point Q on a spherical wavefront at a duration ux and ux from the origins of O and O′ respectively.

For round-trip motion, the mean round-trip pace of light, k, is the same in all inertial frames, so for the point Q with duration vector u = (ux,0, 0) and

ux = ks, and ux′ = ks′.

Consider the dual standard Galilean transformation with a factor λ, which is to be determined and may depend on α = p/k (cf. c/v)

ux = λ(ux − ps) = λ(ux − αks) = λ(ux − αux) = λux(1 − α).

The inverse transformation is the same except that the sign of β is reversed:

ux = λ(ux′ + ps′) = λ(ux′ + αks′) = λ(ux′ + αux′) = λux(1 + α).

Multiply these two equations to get

uxux′ = λ2uxux(1 − α²).

Divide out uxux to get

λ² = 1/(1 − α²), or

λ = 1/√(1 − α²).

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Consider what happens if this approach is applied to the independent distance s times the constant k.

ks′ = ux = λ(ks − αux) = λ(ux − αux) = λux(1 − α).

The inverse transformation is the same except that the sign of α is reversed:

ks = ux = λ(ks′ + αux) = λ(ux′ + αux) = λux(1 + α).

These are exactly the same equations as before, so the result will be the same.

RG 2026-04-04