Showing that Klein-Gordon wave equation is relativistic?

Question

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 $x=0$?

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.