Coupled Oscillator's Stiffness and speed of light

Question

In Schwabl book (Advanced Quantum Mechanics) page 258, in his triumph to show the relation between the coupled oscillators and Klein-Gordon equation he finds the following relation which is the equation of motion for a 1-d coupled oscillators on a discrete lattice: $$\partial^2_tq(X,t)-\nu^2 \nabla^2 q(X,t)+\Omega_0^2 q(X,t)=0$$ where $\Omega_0$ is the harmonic potential of individual particles and $\nu$ is stiffness constant. and $q(X,t)$ is the displacement from the equilibrium position of each lattice point in the direction of vector $X$.

afterward, he introduces the following substitution:

$\nu \to c$ , $\frac{\Omega_0}{\nu} \to m$ , $(X,t)=x$ , and $q(X,t) \to \phi(x)$.

where he claims by doing the above substitution one obtain the Klein-Gordon equation i.e. $$\partial_\mu\partial^u \phi(x)+m^2\phi(x)=0$$.

Now I see how these substitutions lead to the K.G equation.

My question is why he substitutes stiffness constant ($\nu$) by the speed of light (c)?

Answer 1

First I re-write your equation as $$ \partial_{0}^2 \phi(x) - v^2 \sum_{j=1}^3 \partial_{j}^2 \phi(x) + \Omega_0^2 \phi(x) = 0 $$ Plugging in $v=c$ and $\Omega_0 = mc$ your equation is $$ \partial_{0}^2 \phi(x) - c^2 \sum_{j=1}^3 \partial_{j}^2 \phi(x) + m^2 c^2 \phi(x) = 0 $$ Divide by $c^2$ and you get $$ \frac{1}{c^2}\partial_{0}^2 \phi(x) - \sum_{j=1}^3 \partial_{j}^2 \phi(x) + m^2 \phi(x) = 0 $$

Which you can write in terms of the d'Alembertian $\partial_{\mu} \partial^{\mu} = \frac{1}{c^2}\partial_{t}^2 - \sum_{j=1}^3 \partial_{j}^2$, and you get the Klein-Gordon equation, as claimed.