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# Lorentz transformation derivations

What follows are four derivations of the Lorentz transformation from the complete Galilei (Galilean) transformations in space with time (3+1) and time with space (1+3). Their intersection is linear space and time (1+1), which is the focus of the derivations. The other dimensions may be reached by rotations in space or time.

I. Space with Time (3+1)

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 distance 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, is the same in both frames, so for the point P:

x = ct, and x′ = ct′.

A. Time velocity

Define velocity v as the time velocity vt = ds/dt. Consider the standard Galilean transformation of ct = x with 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 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).

B. Space velocity

Define velocity v as the space velocity vs = (dt/ds)−1. Consider the transposed standard Galilean transformation of ct = x with a factor γ, which is to be determined and may depend on β, where β = v/c:

ct′ = γ(ct − βx) = γct(1 − β).

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

ct = γ(ct′ + βx′) = γct′(1 + β).

Multiply these two equations to get

c²tt′ = γ²c²tt′(1 − β²).

Divide out c²tt′ to get

γ² = 1/(1 − β²),

or γ = 1/√(1 − β²).

Therefore, the Lorentz transformation for space with time is

ct′ = γ(ct − βx) and x′ = γ(x − βct) with γ = 1/√(1 − β²).

II. Time with Space (1+3)

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 distime t and t′ from the origins of O and O′ respectively. According to the second postulate of the special theory of relativity the pace of light, k, is the same in both frames, so for the time-point Q:

t = kx, and t′ = kx′.

A. Space lenticity

Define lenticity w as the space lenticity ws = dt/ds. Consider the dual standard Galilean transformation of t/k = x with a factor λ, which is to be determined, which may depend on β = k/w:

kx’ = t′ = γ(xt/w) = γ(xβt/k) = λt(1 − β).

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

t = λ(x + t’/w) = λ(t’ + βx/k) = λt’(1 + β).

Multiply these two equations to get

tt′ = λ²tt′(1 − β²).

Divide out tt′ to get

λ² = 1/(1 − β²) = γ², or

λ = 1/√(1 − β²) = γ.

B. Time lenticity

Define lenticity w as the time lenticity wt = (ds/dt)−1. Consider the transposed dual standard Galilean transformation of t = kx with a factor λ, which is to be determined and may depend on β = k/w:

t’ = kx′ = λ(kxtk/w) = λ(kx − βt) = λkx(1 − β).

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

kx = λ(kx’ + t’k/w) = λ(kx’ + βt’) = λkx’(1 + β).

Multiply these two equations to get

k²xx′ = λ²k²xx′(1 − β²).

Divide out k²xx′ to get

λ² = 1/(1 − β²) = γ², or

λ = 1/√(1 − β²) = γ.

Therefore, the dual Lorentz transformation for time with space is

x′ = γ(xβt/k) = γ(xt/w),

t′ = γ(kxβt) = γ(kxtk/w).

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

Revised 2021-06-24.