# Lorentz transformation derivations

What follows are four derivations of the Lorentz transformation from the complete Galilei (Galilean) transformations for the time domain and the distance domain. These derivations focus on one dimension. Other dimensions may be reached by rotations of length or duration.

I. The 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 x and x′ from the origins of O and O′ respectively. According to the second postulate of the special theory of relativity the speed of light, c = κ−1, is the same in both frames, so for the point P:

x = ct, and x′ = ct′ (also κx = t, and κx′ = t′).

A. Velocity (cf. inverse lenticity)
Define the velocity v as the time rate of displacement v = ds/dt. Consider the standard Galilean transformation with x = ct and a factor γ, which is to be determined and may depend on β, where β = v/c:

x′ = γ(x − vt) = γ(x − βct) = γx(1 − β).

The inverse transformation is the same except that the primes are interchanged and the sign of β is reversed:

x = γ(x′ + vt′) = γ(x + βct) = γx′(1 + β).

Multiply these two equations to get

xx′ = γ²xx′(1 − β²).

Divide out xx′ to get

γ² = 1/(1 − β²), or γ = 1/√(1 − β²).

Therefore, the Lorentz transformation for the time domain is

x′ = γ(xvt), and

ct′ = γ(ctβx).

B. Inverse velocity (cf. lenticity)
Define the inverse velocity as w = v−1 = (dx/dt)−1. Consider the transposed dual standard Galilean transformation with κ = c−1, t = κx and a factor λ, which is to be determined and may depend on α = w/κ:

t′ = λ(twx) = λ(tαt) = λt(1 − α).

The inverse transformation is the same except that the primes are interchanged and the sign of α is reversed:

t′ = λ(t + wx) = λ(t + αt) = λt(1 + α).

Multiply these two equations to get

tt′ = λ²tt′(1 − α²).

Divide out tt′ to get

λ² = 1/(1 − α²), or

λ = 1/√(1 − α²).

Therefore, the dual Lorentz transformation for the distance domain with inverse velocity is

κx′ = λκx(1 − α), and

t′/κ = λ(t/κ − αx).

Compare ct′ = γ(ctβx) and x′ = γ(xβct). Duality is equivalent to interchanging cκ and xt.

II. The Distance Domain

Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with lenticity w with respect to O in the positive t-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 time-point Q on a spherical wavefront at a length z and z′ from the origins of O and O′ respectively. According to the second postulate of the special theory of relativity the pace of light, κ, is the same in both frames, so for the time-point Q:

z = κs, and z′ = κs′.

A. Lenticity (cf. inverse velocity)

Define the lenticity w as the distance rate of dischronment w = dz/ds. Consider the dual standard Galilean transformation with z = κs and a factor λ, which is to be determined, which may depend on α = w/κ:

z′ = λ(zws) = λ(zαz) = λz(1 − α).

The inverse transformation is the same except that the primes are interchanged and the sign of α is reversed:

z = λ(z′ + ws′) = λ(z′ + αz′) = λz′(1 + α).

Multiply these two equations to get

zz′ = λ²zz′(1 − α²).

Divide out zz′ to get

λ² = 1/(1 − α²), or

λ = 1/√(1 − α²), cf. γ above.

Therefore, the dual Lorentz transformation for the distance domain is

z′/κ = λ(z/κ − αs) and

κs′ = λ(κsαz).

B. Inverse lenticity (cf. velocity)

Define the inverse lenticity as u = w−1 = (dz/ds)−1. Consider the transposed dual standard Galilean transformation with c = κ−1, z = κs and a factor γ, which is to be determined and may depend on β = κ/w = u/c:

s′ = γ(suz) = γ(sβs) = γs(1 − β).

The inverse transformation is the same except that the primes are interchanged and the sign of β is reversed:

s′ = γ(s + uz) = γ(s + βs) = γs(1 + β).

Multiply these two equations to get

ss′ = γ²ss′(1 − β²).

Divide out ss′ to get

γ² = 1/(1 − β²), or γ = 1/√(1 − β²).

Therefore, the Lorentz transformation for the distance domain with inverse lenticity is

s′ = γ(suz) and

cz′ = γ(czβs).

Revised 2021-06-24, 2024-04-23, 2024-07-20, and 2024-07-23.