Deriving the boost from symmetry

A change of inertial coordinates must not care where or when it is applied — that is what homogeneity of space and time means. If the coefficients of the map depended on x or t, two identical rods would transform differently at different places and the frame would have a preferred location. So the coefficients are constants and the map is affine; letting the origins coincide at t = t′ = 0 removes the constant terms and leaves it linear. The same conclusion arrives from dynamics: a free particle draws a straight line in (x, t) in every inertial frame, and a map defined on all of spacetime that carries every straight line to a straight line is affine — the fractional-linear maps that also preserve lines blow up on some hyperplane, which homogeneity forbids. Now track the origin of the moving frame. The event x′ = 0 must happen at x = vt, which forces the first row to read x′ = a(v)(x − vt). The second row is still wide open, t′ = b(v)t + d(v)x, and nothing about light has been used.

Reversing the roles of the two frames must undo the boost: the inverse of a boost by v is a boost by −v. Reflect space, x → −x, and v → −v with it, so a and b are even functions of v while d, which multiplies a length inside a time, is odd. Now invert the matrix explicitly. With Δ = det = a(b + vd), the inverse gives x = (b x′ + a v t′)/Δ and t = (a t′ − d x′)/Δ, while the −v boost reads x = a(x′ + vt′) and t = bt′ − dx′. Match coefficient by coefficient. The t′ term in x demands av = av/Δ, so Δ = 1 outright; the x′ term then demands a = b ≡ γ(v). The unit determinant is forced, not assumed, and no sign choice was available. Then γ(γ + vd) = 1 fixes d = (1 − γ²)/(γv), which is tidier written as d = −γKv with K ≡ (1 − γ⁻²)/v². Exactly one unknown function of v is left.

Write the boost as a matrix acting on the column (t, x): M(v) = γ[[1, −Kv], [−v, 1]]. Let K still depend on v, and compose two boosts, v₁ then v₂. The product has diagonal entries 1 + K₂v₁v₂ and 1 + K₁v₁v₂. But a boost matrix has equal diagonal entries, so if the product is to be a boost at all — group closure, without which the relativity principle is empty — then K₁ = K₂ for every pair of speeds. K is one universal constant with units s² m⁻². The off-diagonal entries then hand over v₃ = (v₁ + v₂)/(1 + Kv₁v₂) and γ₃ = γ₁γ₂(1 + Kv₁v₂). Feed the now-constant K back into K = (1 − γ⁻²)/v² and solve: γ = (1 − Kv²)^(−1/2). The entire transformation is written, and exactly one number in it is unknown. A note on names: some accounts call the invariant speed itself K, so that Galileo is K → ∞; here K is the coefficient multiplying vx in the time row, the invariant speed is K^(−1/2), and Galileo is K → 0.

Three worlds are left, one per sign of K. K = 0 gives γ = 1 identically, t′ = t and x′ = x − vt: Galileo, exactly, with velocities simply adding. K < 0 is the Euclidean branch; write K = −1/κ², so γ = (1 + v²/κ²)^(−1/2) is less than 1 and moving clocks run fast. Set v = κ tan θ and boosts compose by adding θ, so two boosts of κ each reach θ = 90° and an infinite velocity, and larger angles flip the sign. Worse, for a signal with dx = u dt and u > 0, the transformed interval dt′ = γ dt(1 + uv/κ²) turns negative for any v more negative than −κ²/u; since v is unbounded on this branch such a frame always exists, so every cause-and-effect pair can be reversed. That is the exclusion — causality, not experiment. K > 0 survives, carrying exactly one speed V = K^(−1/2) that every frame agrees on.

On the surviving branch symmetry has produced an invariant speed but not its value. Any experiment returning a finite invariant speed sets K. The second postulate is the historically convenient one: light travels at the invariant speed, so V = c = 2.998 × 10⁸ m s⁻¹ and K = 1/c² = 1.113 × 10⁻¹⁷ s² m⁻². Note carefully what this does and does not claim. It does not make relativity a theory about light. If the photon carried a small mass, light would travel a shade below the invariant speed, c would still be the invariant speed, and every formula in this unit would stand unchanged — only the road in would need repaving. Modern Ives–Stilwell tests on stored ion beams bound any deviation from the derived γ at a few parts in 10⁸, and what they measure is K, through the rate of a moving clock, not the speed of light.

The argument is cheap but not free. It assumes flat spacetime and globally defined inertial coordinates, so gravity — which makes homogeneity fail from place to place — sits outside it, and that is where general relativity begins. It assumes isotropy, which is what allowed γ to be even and what forces the transverse coordinates to go untouched, y′ = y and z′ = z. And it delivers one boost direction: the full symmetry needs rotations added to make the Lorentz group and translations added to make the ten-parameter Poincaré group. What it cannot do is name K, choose a branch without a causality axiom, or say anything whatever about dynamics. Momentum, mass–energy and the transformation of fields are separate arguments built on top of this kinematics, not consequences of it.