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(x − t/w) = γ(x − βt/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(kx − tk/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(x − βt/k) = γ(x − t/w),
t′ = γ(kx − βt) = γ(kx − tk/w).
Compare ct′ = γ(ct − βx) and x′ = γ(x − βct). Duality is equivalent to interchanging c ↔ k and x ↔ t.
Revised 2021-06-24.