Using normalized constants, the classical wave equation

\begin{equation} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2} \end{equation}

differs from the Klein-Gordon wave equation by

\begin{equation} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2} - m^2\psi \end{equation}

The Klein-Gordon [2] wave equation is also said to be relativistic. This implies that the classical [1] wave equation is not relativistic.

What is meant by a "relativistic" wave, and what specifically can we measure from the wave itself that tells whether this wave is relativistic or not?

Is there a specific motion path in 1D for which the classical wave [1] does not appear the same as for the stationary observer's reference frame at $x=0$?

  • 2
    $\begingroup$ Any classical wave equation where the wave velocity is $c$ (and you don't have weird non-invariant extra terms) is Lorentz-invariant, and therefore "relativistic" in my book. $\endgroup$ Commented Dec 9, 2022 at 20:20
  • 1
    $\begingroup$ This implies… What is your logic for concluding that? Do you understand that “being relativistic” means “being form-invariant under Lorentz transformations”? Both equations have this property. $\endgroup$
    – Ghoster
    Commented Dec 9, 2022 at 21:29
  • $\begingroup$ Your final paragraph implies that you think solutions of relativistic wave equations look the same to all inertial observers. This is not the case. The Doppler shift of light is an example. Observers with relative motion observe a different frequency and wavelength when observing the same light wave. $\endgroup$
    – Ghoster
    Commented Dec 9, 2022 at 21:50
  • 1
    $\begingroup$ I recommend using the Lorentz transformation of coordinates to work out how the combination of second derivatives in both equations transforms. $\endgroup$
    – Ghoster
    Commented Dec 9, 2022 at 21:57
  • $\begingroup$ @Ghoster thank you. it seems i misunderstood the comparison is to Schrodinger wave equation, which is non-relativistic, not to the classical wave equation. $\endgroup$
    – James
    Commented Dec 10, 2022 at 3:45

1 Answer 1


Let's take the "intuitive" derivation from quantum mechanics books: $$ E^2=\mathbf{p}^2c^2 + m^2c^4 $$ Quantization is $$ E\rightarrow i\hbar\frac{\partial}{\partial t},\mathbf{p}\rightarrow-i\hbar\nabla,$$ so the energy-momentum relation becomes $$ -\hbar^2\frac{\partial^2\psi}{\partial t^2}=-\hbar^2\nabla^2\psi + m^2c^4\psi \leftrightarrow \frac{\partial^2\psi}{\partial t^2}-\nabla^2\psi + \frac{m^2c^4}{\hbar^2}\psi $$

If we performed the same operations with non-relativistic dispersion relation $$E=\frac{\mathbf{p}^2}{2m},$$ we would get the Schrödinger equation for free particle.

  • $\begingroup$ so the energy is either correct/incorrect relativistically. Is there also something physically/observably wrong when a wave is not relativistic? $\endgroup$
    – James
    Commented Dec 9, 2022 at 20:24
  • $\begingroup$ @James on more serious level, Klein Gordon is invariant to lorentz transformations, whereas Schrödinger is not - all the usual differences between relativistic and not are there. $\endgroup$
    – Roger V.
    Commented Dec 9, 2022 at 20:52
  • $\begingroup$ This is probably a stupid question considering I have no knowledge about the subject but, how do squares in eq 1 become second-order derivatives in eq 3? $\endgroup$ Commented Dec 9, 2022 at 20:56
  • 2
    $\begingroup$ @GedankenExperimentalist Squaring an operator means applying it twice. The operator here is “take a derivative”. Applying this twice gives a second derivative. $\endgroup$
    – Ghoster
    Commented Dec 9, 2022 at 21:37
  • $\begingroup$ @Ghoster like taking the divergence of a gradient? $\endgroup$ Commented Dec 9, 2022 at 21:40

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