Is $ \partial_{\mu} \partial^{\mu} $ the second derivative or derivative squared? 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 $
 A: I think a quick brush up on the index notation and relativity will clear this up. (In what follows I'll be using units in which $c=1$.) First consider the operator $\partial_\mu$, which is the four-gradient, represented by: $$\partial_\mu \equiv \begin{pmatrix} \partial_t & \partial_x & \partial_y& \partial_z\end{pmatrix}.$$
This is an operator since it can act on a Lorentz scalar and produce a four-vector, just like the "normal" gradient operator can act on a scalar field and produce a vector field. So $\partial_\mu \Phi$ we mean the object:
$$\partial_\mu \Phi \equiv \begin{pmatrix}\partial_t\Phi & \partial_x\Phi & \partial_y\Phi & \partial_z\Phi \end{pmatrix},$$
and therefore the quantity $\partial^\mu \Phi \partial_\mu \Phi$ is just shorthand for:
$$\partial^\mu \Phi \partial_\mu \Phi = \eta^{\mu\nu} \partial_\nu \Phi \partial_\mu \Phi = -(\partial_t \Phi)^2 + (\partial_x\Phi)^2 + (\partial_y\Phi)^2 + (\partial_z\Phi)^2.$$
where I've used the $(- + + +)$ metric signature.
However, the quantity $\partial_\mu \partial^\mu$ is another beast altogether: if you follow the index notation conventions, then: $$\partial_\mu \partial^\mu \equiv \eta^{\mu\nu}\partial_\nu \partial_\mu = - \left(\frac{\partial}{\partial t}\right)^2 + \left(\frac{\partial}{\partial x}\right)^2 + \left(\frac{\partial}{\partial y}\right)^2 + \left(\frac{\partial}{\partial z}\right)^2.$$
Such an operator can also act on a scalar field, but it produces a Lorentz scalar (very much like the Laplacian $\nabla^2$ can act on a scalar and produce a scalar), and so $$\partial_\mu \partial^\mu \Phi = - \frac{\partial^2 \Phi}{\partial t^2} + \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} + \frac{\partial^2 \Phi}{\partial z^2}.$$
I'll leave it to you to see why these two quantities aren't the same.
A: Using a $(+1,-1,-1,-1)$ Minkowski metric,
$$\Box\Phi=\partial_\mu\partial^\mu\Phi=\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}-\frac{\partial^2\Phi}{\partial x^2}- \frac{\partial^2\Phi}{\partial y^2}-\frac{\partial^2\Phi}{\partial z^2}$$
while
$$\partial_\mu\Phi\partial^\mu\Phi=\frac{1}{c^2}\left(\frac{\partial\Phi}{\partial t}\right)^2-\left(\frac{\partial\Phi}{\partial x}\right)^2– \left(\frac{\partial\Phi}{\partial y}\right)^2 -\left(\frac{\partial\Phi}{\partial z}\right)^2.$$
The operator $\partial_\mu\partial^\mu$ by itself always means
$$\partial_\mu\partial^\mu=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}-\frac{\partial^2}{\partial z^2}.$$
But note that this can also be written as
$$\partial_\mu\partial^\mu=\left(\frac1c\frac{\partial}{\partial t}\right)^2-\left(\frac{\partial}{\partial x}\right)^2– \left(\frac{\partial}{\partial y}\right)^2 -\left(\frac{\partial}{\partial z}\right)^2.$$
A second derivative as an operator is a repeated (“squared”) derivative. But a second derivative of $\Phi$ is not the square of the first derivative of $\Phi$. The former is linear in $\Phi$ while the latter is quadratic in $\Phi$.
