The boost in standard configuration and matrix form

The transformation is usually first met as t′ = γ(t − vx/c²) and x′ = γ(x − vt). Written that way its coefficient matrix is [[γ, −γv/c²], [−γv, γ]]: the entries carry three different units, and the matrix is not symmetric. Replace the time coordinate by ct, measured in metres exactly like x, and the same content reads ct′ = γ(ct − βx) and x′ = γ(x − βct) with β = v/c. Every entry of Λ = [[γ, −γβ], [−γβ, γ]] is now a pure number, the matrix is symmetric, and interchanging ct with x interchanges the two equations. The conversion is easy to carry in your head: one metre of ct is the 3.3 ns light takes to cross it, so 1.00 μs is 300 m. Nothing physical has changed — this is bookkeeping — but it is why a boost ends up looking like a rotation rather than like an untidy pair of formulas.

Four conditions are packed into that phrase, and the derivation uses all four. S′ moves at constant velocity +v along the common x-axis; the two sets of axes are parallel; the origins coincide, so the event (0, 0, 0, 0) in S is the event (0, 0, 0, 0) in S′, which is what allows a linear map with no additive constant; and the clocks of each frame are Einstein-synchronised, because t is not a number until a synchronisation convention has fixed it. The transverse coordinates pass straight through, y′ = y and z′ = z, and that is a symmetry result rather than an omission: if a boost changed transverse lengths, two identical rods passing each other sideways would each have to be shorter than the other. Lose a condition and you lose the matrix. Origins that do not coincide add a translation, which is the Poincaré group and not a boost; a relative velocity that is not along a coordinate axis has to be handled by rotating into standard configuration, boosting, and rotating back.

Take β = 0.8, so γ = 1/√(1 − 0.64) = 1/0.6 = 5/3. An event is recorded in S at ct = 500 m, x = 400 m. Multiply: ct′ = (5/3)(500 − 0.8 × 400) = (5/3)(180) = 300 m, and x′ = (5/3)(400 − 0.8 × 500) = 0. The event sits at the spatial origin of S′, which is no accident — x = βct = 0.8 × 500 = 400 m is exactly the worldline that origin traces in S, and that worldline is the ct′ axis. Notice also that 300 m = 500 m ÷ γ. That is time dilation, but it showed up only because this event happens to satisfy x′ = 0; the matrix by itself never says t′ = t/γ. Check the arithmetic with the invariant: (ct)² − x² = 250 000 − 160 000 = 90 000 m², and (ct′)² − x′² = 90 000 − 0 = 90 000 m². Check every transformed pair this way — the interval is the cheapest error trap you have.

Solve the two lines for ct and x, or simply multiply: Λ(β)Λ(−β) has diagonal entries γ² − γ²β² = 1 and off-diagonal entries γ²β − γ²β = 0, so Λ(β)⁻¹ = Λ(−β), giving ct = γ(ct′ + βx′) and x = γ(x′ + βct′). Because γ is built from β², undoing a boost costs nothing but a pair of sign changes, and that is reciprocity turned into an identity: if S says S′ moves at +v, S′ says S moves at −v, and one γ covers both statements. Two warnings travel with it. Inverting is not transposing — Λ is already symmetric here, so Λᵀ = Λ, while Λ⁻¹ differs from Λ at every β other than zero. And the two signs have to flip together: pairing ct = γ(ct′ + βx′) with x = γ(x′ − βct′) mixes the two directions, and the resulting matrix has determinant γ²(1 + β²), which is 2.125 at β = 0.6. It is still an invertible linear map, but it is not a Lorentz transformation, because it does not leave (ct)² − x² alone.

det Λ = γ² − (γβ)² = γ²(1 − β²) = 1 for every β < 1, which says a region of the (ct, x) plane keeps its area: the boost shears a square of events into a parallelogram of the same area. Useful, and cheap to check — but it does not identify a boost. Write the Galilean transformation in these same coordinates, ct′ = ct and x′ = x − βct, and its matrix [[1, 0], [−β, 1]] has determinant exactly 1 at any β whatsoever. The condition that does the identifying is ΛᵀηΛ = η with η = diag(1, −1), which is nothing but the demand that (ct′)² − x′² = (ct)² − x² for every event; a Euclidean rotation satisfies RᵀR = I instead, and that single minus sign inside η is the entire difference between turning a plane and boosting a frame. Apply the condition to the Galilean shear and you get [[1 − β², β], [β, −1]], which equals η only at β = 0. Apply it to Λ(β) and the γ² and γ²β² terms cancel exactly, returning η. So check the determinant to catch arithmetic slips, and check the interval to know you have a boost.

Feed the matrix a light ray. The column (1, 1), meaning ct = x, comes back as γ(1 − β)(1, 1): the direction survives untouched and only the scale changes, by γ(1 − β) = √((1 − β)/(1 + β)), the longitudinal Doppler factor. The column (1, −1) is stretched by the reciprocal √((1 + β)/(1 − β)), and the two eigenvalues multiply to the determinant, 1; at β = 0.6 they are 0.5 and 2.0. That the two light-line directions are eigenvectors is the second postulate written in linear algebra. Now let β → 0. With γ = 1 + ½β² + …, a speed of 300 m s⁻¹ gives β = 1.0 × 10⁻⁶ and γ − 1 ≈ 5 × 10⁻¹³, so x′ = x − vt is right to roughly twelve figures. The piece that survives longest is the one Galileo never had, t′ = t − vx/c², which at that same speed across x = 1000 km amounts to 3.3 ns — small, but not small enough for a satellite navigation system to ignore.