# Showing that Klein-Gordon wave equation is relativistic?

Using normalized constants, the classical wave equation

$$$$\frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2}$$$$

differs from the Klein-Gordon wave equation by

$$$$\frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2} - m^2\psi$$$$

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 $$x=0$$?

• 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. Commented Dec 9, 2022 at 20:20
• 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. Commented Dec 9, 2022 at 21:29
• 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. Commented Dec 9, 2022 at 21:50
• I recommend using the Lorentz transformation of coordinates to work out how the combination of second derivatives in both equations transforms. Commented Dec 9, 2022 at 21:57
• @Ghoster thank you. it seems i misunderstood the comparison is to Schrodinger wave equation, which is non-relativistic, not to the classical wave equation. Commented Dec 10, 2022 at 3:45

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.