The Lorentz transformation

Before any physics, ask what kind of map can relate two inertial frames at all. Space and time are homogeneous: no point and no instant is special, so the rule that converts a displacement (Δx, Δt) into (Δx′, Δt′) cannot depend on where or when the displacement starts. The only maps with that property are linear ones with constant coefficients, x′ = Ax + Bt and t′ = Dx + Et, and the first postulate agrees, since a linear map carries every straight worldline (free motion) into another straight worldline. Transverse coordinates are untouched, y′ = y and z′ = z, by the symmetry argument that also protects the light clock's mirror spacing. Two facts about the S′ origin fix half the coefficients: it sits at x′ = 0 and moves at x = vt in S, so x′ must vanish whenever x = vt, which forces x′ = A(x − vt). The principle of relativity then says that S, seen from S′, is a frame moving at −v and nothing else, so the inverse map has the same shape with the same coefficient: x = A(x′ + vt′). Galileo is the special case A = 1; everything now rests on finding A.

Fire a light pulse from the coincident origins at t = t′ = 0. The second postulate says it obeys x = ct in S and x′ = ct′ in S′: the same pulse, two descriptions. Feed x = ct into x′ = A(x − vt): ct′ = A(c − v)t. Feed x′ = ct′ into the inverse: ct = A(c + v)t′. Multiply the two: c²tt′ = A²(c² − v²)tt′, so A² = c²/(c² − v²) and A = 1/√(1 − v²/c²) = γ. At v = 0.600c that is 1/√0.64 = 1.25. The time equation costs one more line. Substitute x′ = γ(x − vt) into x = γ(x′ + vt′) to get x = γ²(x − vt) + γvt′, then solve: t′ = [x − γ²(x − vt)]/(γv) = γ[t − x(γ² − 1)/(γ²v)]. Since (γ² − 1)/γ² = v²/c² (check at 0.600c: 0.5625/1.5625 = 0.36 = β²), this is t′ = γ(t − vx/c²). Nothing was assumed about clocks or rods; the position term in t′ was forced by insisting that one pulse move at c in both frames.

Written in ct and x the two lines are mirror images: x′ = γ(x − βct) and ct′ = γ(ct − βx), with β = v/c. That symmetry is the point. The time equation is not 'time dilation applied to t'; it says that S′ assigns different times to events S calls simultaneous, by −γβΔx/c, and it says so at first order in β. Compare the two corrections for a car at v = 30 m s⁻¹ and a baseline x = 1.00 km: the offset vx/c² = 3.3 × 10⁻¹³ s, while the γ correction over one second, (γ − 1) × 1 s ≈ β²/2 × 1 s = 5 × 10⁻¹⁵ s, is 67 times smaller. So the Galilean limit needs two conditions, β ≪ 1 and vx/c² ≪ t, not one, and clock synchronisation across a continent at ordinary speeds feels the first-order term long before it feels γ. The inverse map follows from relativity alone: swap primed for unprimed and v for −v. Check it by composing the two: x = γ[γ(x − vt) + vγ(t − vx/c²)] = γ²x(1 − v²/c²) = x, as it must.

Because the map is linear, the difference between two events transforms like a single event: Δx′ = γ(Δx − vΔt) and Δt′ = γ(Δt − vΔx/c²), with the inverse pair obtained by v → −v. Now each headline result is a side condition. A clock at rest in S′ ticks twice at the same place, Δx′ = 0, so the inverse time equation gives Δt = γ(Δt′ + vΔx′/c²) = γΔt′: the proper time Δt′ is the shortest, and S reads γ times more. A rod at rest in S′ has proper length Δx′ = L₀; S measures it by marking both ends at one instant, Δt = 0, so Δx′ = γΔx and L = L₀/γ. At 0.600c a 1.00 μs proper interval reads 1.25 μs and a 1.00 m rod reads 0.800 m. Notice what decided which way γ went: not which object was 'moving', but which difference was set to zero, and in which frame. Impose the other condition, Δt′ = 0, on the same rod and you get L₀ back, correctly, because you have then measured it in its own frame.

Put x = ct into the boost: x′ = γ(c − v)t and ct′ = γ(c − v)t, so x′ = ct′; put x = −ct and x′ = −ct′ follows the same way. Both light rays are fixed lines of the map, for every v < c. More is true: expand c²t′² − x′² = γ²[(ct − βx)² − (x − βct)²] = γ²(1 − β²)(c²t² − x²) = c²t² − x². The boost preserves the quadratic form c²t² − x² exactly as a rotation preserves x² + y², with a minus sign that makes the geometry hyperbolic rather than circular. Setting γ = cosh φ and γβ = sinh φ, consistent because cosh²φ − sinh²φ = 1, turns the boost into x′ = x cosh φ − ct sinh φ, ct′ = ct cosh φ − x sinh φ: a rotation through the rapidity φ = atanh β. Two boosts along one axis therefore compose by adding rapidities, and since atanh 0.600 = ln 2 exactly, 0.600c followed by 0.600c is tanh(ln 4) = 15/17 c = 0.882c, with γ = cosh(ln 4) = 17/8. Correspondence holds here too: for β ≪ 1, φ ≈ β and rapidities add like Galilean speeds.

The formulae assume a lot of tidiness: one boost, along x, with the two x axes collinear, y and z axes parallel, and origins coinciding at t = t′ = 0. Relax the last and a constant shift is added to each coordinate, giving the Poincaré transformation; nothing physical changes, because only differences enter. Relax the direction and you must split the position vector into a part along v, which transforms as x does, and a part perpendicular to it, which does not: r′ = r + [(γ − 1)(r ⋅ v)/v² − γt] v and t′ = γ(t − r ⋅ v/c²). Apply two boosts along different directions and the product is not a pure boost but a boost combined with a spatial rotation (the Thomas–Wigner rotation), so boosts alone do not close into a group; boosts and rotations together do. Above all, both frames must be inertial. An accelerating observer can use the boost to the instantaneously comoving inertial frame at each moment, which is how proper time is integrated along a worldline, but cannot stitch those frames into one global coordinate system, and gravitation lies outside special relativity altogether.