This might be a silly question, but I'm just getting my feet wet with field theories. So far I have assumed that $ \partial_{\mu} \Phi\partial^{\mu}\Phi $ means $ (\Phi_t)^2-(\Phi_x)^2-...$ . I thought this, because it made sense to me, that a Lagrangian density has a kinetic term proportional to some sort of velocity squared ($ \Phi_t^2)$, and I hand-waved it to myself that in field theory a spacial derivative squared is similar.
Let's take the Sine-Gordon Lagrangian for example, the way my mentor wrote it down: $$\mathcal{L}_{SG} = \frac{1}{2}\partial_{\mu} \Phi\partial^{\mu}\Phi +\cos(\Phi) $$ when solved for the equations of motion, I believe this yields something like: $$ \Box\Phi+\sin(\Phi)=0$$ where $ \Box$ is just the d 'Alembertian, $ \Box=\partial_{\mu} \partial^{\mu}$, but now we think of it as a second derivative? I'm assuming so, because another way of writing the EoM is: $ \frac{1}{c^2}\Phi_{tt}-\Phi_{xx}+\sin(\Phi)=0 $