What does $\partial\phi$ mean for a scalar field $\phi(\vec{x},t)$?

Question

If I have a scalar field $\phi \equiv \phi(\vec{x},t)$, and the Lagrangian density is $$\mathcal{L} = \frac{1}{2}(\partial\phi)^2 + ...$$ what does $\partial\phi$ mean or in other words how do I expand $\partial\phi$ out?

Answer 1

I believe that $\partial$ has no index because it is being squared. What really is written is

$ (\partial \phi)^2 = \partial_\mu \phi \partial^\mu \phi $

which has no indices. Therefore, the author writes $ \partial $ without indices to indicate that the resulting object is a scalar.

Answer 2

It depends on the context and it should say somewhere in your textbook. Often it is a shorthand for \begin{align}(\partial \phi)^2&=(\partial_\mu\phi)(\partial^\mu\phi)\\ &=\eta^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) \end{align} where $\eta_{\mu\nu}$ is the Minkowski metric. Depending on your textbook the Minkowski metric is either $\text{diag}(-1,+1,+1,+1)$ or $\text{diag}(+1,-1,-1,-1)$. I used brackets to be precise but usually this term is written as $\partial_\mu\phi\partial^\mu\phi$.