It’s a good exercise to check the invariant interval for both subluminal and superluminal objects. Let’s do this with the delta form of the Lorentz transformations:
Subluminal case:
This is a check that c²(Δt´)² – (Δx´)² – (Δy´)² – (Δz´)² = c²(Δt)² – (Δx)² – (Δy)² – (Δz)².
The Lorentz transformation is
cΔt´ = γ (cΔt – vΔx/c), Δx´ = γ (Δx – vΔt), Δy´ = Δy, Δz´= Δz.
So we have
γ² (cΔt – vΔx/c)² – γ² (Δx – vΔt)² – (Δy)² – (Δz)²
= γ² (c²(Δt)² – vΔtΔx + v²(Δx)²/c² – (Δx)² + vΔtΔx – v²(Δt)²) – (Δy)² – (Δz)²
= γ² ((c² – v²)(Δt)² – (1 – v²/c²)(Δx)²) – (Δy)² – (Δz)²
= γ² (1 – v²/c²)(c²(Δt)² – (Δx)²) – (Δy)² – (Δz)²
= c²(Δt)² – (Δx)² – (Δy)² – (Δz)².
Superluminal case:
This is a check that c²(Δr´)² – c²(Δt1´)² – c²(Δt2´)² – c²(Δt3´)² = (Δr)² – c²(Δt1)² – c²(Δt2)² – c²(Δt3)².
The Lorentz transformation is
Δr´ = γ (Δr – c²Δt/v), cΔt1´ = γ (cΔt1 – cΔr/v), cΔt2´ = cΔt2, cΔt3´ = cΔt3.
So we have
γ² (Δr – c²Δt/v)² – γ² (cΔt1 – cΔr/v)² – c²(Δt2)² – c²(Δt3)²
= γ² ((Δr)² – 2c²ΔrΔt1/v + c4(Δt1)²/v² – c²(Δt1)² + 2c²Δt1Δr/v – c²(Δr)²/v²) – c²(Δt2)² – c²(Δt3)²
= γ² ((Δr)²(1 – c²/v²) – c²(Δt1)²(1 – c²/v²)) – c²(Δt2)² – c²(Δt3)²
= (Δr)² – c²(Δt1)² – c²(Δt2)² – c²(Δt3)².