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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 and 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 and 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) = γct(1 − β) and x′ = γ(x − βct) = γx(1 − β) 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 ℓ 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 an instant 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 instant Q:

t = kx, and t′ = kx′.

A. Space lenticity

Define lenticity ℓ as the space lenticity ℓs = dt/ds. Consider the dual standard Galilean transformation of t and kx with a factor λ, which is to be determined and may depend on β, where 1/β = ℓ/k:

kx′ = t′ = λ(t − ℓx) = λ(tkx/β) = λt(1 − 1/β).

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

t = λ(t + ℓx) = λ(t′ + kx′/β) = λt′(1 + 1/β).

Multiply these two equations to get

tt′ = λ²tt′(1 − 1/β²).

Divide out tt′ to get

λ² = 1/(1 − 1/β²) = β²/(β² − 1) = −β²γ², or

λ = 1/√(1 − 1/β²) = β/√(β² − 1) = −βγ.

B. Time lenticity

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

t′ = kx′ = λ(kx − ℓt/k) = λ(kxt/β) = λkx(1 − 1/β).

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

kx = λ(kx + ℓt/k) = λ(kx′ + t′/β) = λkx′(1 + 1/β).

Multiply these two equations to get

k²xx′ = λ²k²xx′(1 − 1/β²).

Divide out k²xx′ to get

λ² = 1/(1 − 1/β²) = β²/(β² − 1) = −β²γ², or

λ = 1/√(1 − 1/β²) = β/√(β² − 1) = −βγ.

Therefore, the dual Lorentz transformation for time with space is

t′ = kx′ = −βγ(t − ℓx) = −βγ(tkx/β) = −βγt(1 − 1/β) = −γ(βtkx) = γ(kxβt) = γt(1 − β)

kx′ = t′ = −βγ(kx − ℓt/k) = −βγ(kxt/β) = −βγkx(1 − 1/β) = γ(tβkx) = γkx(1 − β).

cf. ct′ = γ(ctβx) and x′ = γ(xβct). In short, it is equivalent to interchanging ck and xt.