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

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?

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.
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 $$\partial_\mu\phi\partial^\mu\phi$$.