Does the wave equation keep its form?

A light wave obeys a single wave equation. Describe the same wave from a moving frame with the everyday change of coordinates and the equation no longer holds in its old form; use the change of coordinates Einstein derived, and it does.

Take a wave travelling at c along x, φ = cos(k(x − ct)), which satisfies ∂²φ/∂x² − (1/c²)∂²φ/∂t² = 0. Describe it from a frame moving at v. With the Galilean substitution x′ = x − vt, t′ = t, the wave no longer satisfies an equation of the old form in the new coordinates: it leaves a residual of relative size |β(2 − β)|, 0.19 at β = 0.1, and the correct operator there is a different one, (1 − β²)∂²/∂x′² + (2v/c²)∂²/∂x′∂t′ − (1/c²)∂²/∂t′². With the transformation of §3 of the relativity paper the equation keeps its form and the residual is 0 at every speed. Section 6 shows the same for the full Maxwell–Hertz equations of empty space, with the fields transforming as well; this instrument checks only the scalar wave operator.

In the moving frame x = x′ + vt′ and t = t′, so x − ct = x′ − (c − v)t′, and the wave is cos(k(x′ − (c − v)t′)): it moves at c − v. Differentiating twice in x′ gives −k²φ, and twice in t′ gives −k²(c − v)²φ. The old-form operator then gives −k²φ + (k²(c − v)²/c²)φ = −k²φ[1 − (1 − β)²] = −k²φβ(2 − β). At β = 0.1, β(2 − β) = 0.1 × 1.9 = 0.19, and since φ is at most 1 the largest residual is 0.19k². With the relativity transformation, x = γ(x′ + vt′) and t = γ(t′ + vx′/c²), so x − ct = γ(1 − β)(x′ − ct′). The wave is cos(k′(x′ − ct′)) with k′ = γ(1 − β)k: it still moves at c, and the old-form operator gives −k′²φ + (k′²c²/c²)φ = 0. The wavenumber changes, by the Doppler factor γ(1 − β) = √((1 − β)/(1 + β)) = 0.905 at β = 0.1, but the equation does not.