# Invariant intervals

Let’s begin with the space-time invariant interval r´² − ct´² = r² − ct². Then let us solve the equations:

r´ = Ar + Bt and t´ = Cr + Dt.

r´ = 0 = Ar + Btr = −tB/A = vt where v = −B/A {or} B = −Av

r´ = Ar + Bt = A(rvt)

A²(rvt)² − c²(Cr + Dt)² = r² − c²t²

A²r² − 2A²vrt + A²v²t² − C²c²r² − 2CDc²rtD²c²t² = r² − c²t²

⇒ (A² − C²c²)r² = r² {or} A² − c²C² = 1

⇒ (A²v² − D²c²)t² = −c²t² {or} D²c² − A²v² = c²

⇒ (2A²v + 2CDc²)rt = 0 {or} CDc² = −A²v

C²c² = A² − 1

⇒  D²c² = c² + A²v²

C²D²c²c² = (A² − 1)(c² + A²v²) = A²A²v²

A²c² − c² + A²A²v² − A²v² = A²A²v²

A²c² − A²v² = c²

A²(c² − v²) = c²

A² = c²/(c² − v²)

A² = 1/(1 − v²/c²) = γ²

B = −

C²c² = γ² − 1

C²c² = (v²/c²)γ²

C = −(v/c²)γ

D²c² = c² + v²γ² = c²γ²

D² = γ² {or} D = γ

We have

A = γ

B = −γv

C = −γv/c²

D = γ

The Lorentz transformations are then

r´ = γ(rvt)

t´ = γ(trv/c²)

Let’s begin with the space-time invariant interval t´² − k²r´² = t² − k²t² with k = pace of light. Then let us solve the equations:

t´ = At + Br and r´ = Ct + Dr.

t´ = 0 = At + Brt = −rB/A = ur where u = −B/A {or} B = −Au

t´ = At + Br = A(tur)

A²(tur)² − k²(Ct + Dr)² = t² − k²r²

A²t² − 2A²urt + A²u²r² − C²k²t² − 2CDk²rtD²k²r² = t² − k²r²

⇒ (A² − C²k²)t² = t² {or} A² − C²k² = 1

⇒ (A²u² − D²k²)r² = −k²r² {or} D²k² − A²u² = k²

⇒ (2A²u + 2CDk²)rt = 0 {or} CDk² = −A²u

C²k² = A² − 1

⇒  D²k² = k² + A²u²

C²D²k²k² = (A² − 1)(k² + A²u²) = A²A²u²

A²k² − k² + A²A²u² − A²u² = A²A²u²

A²k² − A²u² = k²

A²(k² − u²) = k²

A² = k²/(k² − u²)

A² = 1/(1 − u²/k²) = μ²

B = −

C²k² = μ² − 1

C²c² = (v²/c²)γ²

C = −(u/k²)μ

D²k² = k² + u²μ² = k²μ²

D² = μ² {or} Dμ

We have

Aμ

B = −μu

C = −μu/k²

Dμ

The Lorentz transformations are then

t´ = μ(tur)

r´ = μ(rtu/k²)