# Lagrange density for massless scalar field [duplicate]

I am reading a book on QFT which is stating the following.

For a massless scalar field $\phi$ the simplest possible Lagrangian is given by $$\mathcal{L}(x) = \frac{1}{2} \partial^\mu\phi \partial_\mu\phi$$ with $\partial_\mu\phi\equiv\partial\phi(x)/ \partial\phi^\mu$. This can be expanded to $\mathcal{L}(x) = \frac{1}{2} (\partial_t\phi)^2 -\frac{1}{2}\nabla\phi\cdot\nabla\phi$. Which I easily see by using the definitions of $\partial^\mu$ and $\partial_\mu$ and having the mixted termes cancelling out each other.

But now the book also states that $$\mathcal{L}(x) = \frac{1}{2} \partial^\mu\phi \partial_\mu\phi=\frac{1}{2}(\partial_\mu\phi)^2,$$ but I totaly fail to see this relation. To my understanding the expantion of the right part should look like $$\frac{1}{2}(\partial_\mu\phi)^2= \frac{1}{2} (\partial_t\phi)^2 -\frac{1}{2}\nabla\phi\cdot\nabla\phi+ \partial_t\phi\nabla\phi$$ which is not equal to the given expantion above.

So what is my error?

 Thanks for the comments that $$(\partial_\mu\phi)^2 \equiv (\partial_{\mu}\phi)g^{\mu\nu}(\partial_{\nu}\phi)$$ And that I should not take the square "seriously". But If I don't take it seriously, how can I later on see that $$\partial_\mu \left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} \right)= \partial_\mu\partial^\mu\phi$$ when using the Euler-Lagrange equation?

## marked as duplicate by AccidentalFourierTransform, Kyle Kanos, peterh, ZeroTheHero, WolpertingerMay 14 '17 at 13:34

You don't have to take that square "seriously". I mean, it's serious, but it is a notation. Remember that $\partial_\mu\phi$ is a vector, so you can't use the same old rule of the square of scalar quantities (with the double product). As there is no possibility of confusion, look at it that way: every time you see the square of a vector, what it really means is $$(V^\mu)^2=V^\mu V_\mu,$$ that is, you take this different sort of square and you recover the same old lagrangian. Remember that indices are raised and lowered through the use of a metric $g_{\mu\nu}$ (that in field theory is usually the Minkowski metric), so you can write the previous as $$(V^\mu)^2=g_{\mu\nu}V^\mu V^\nu.$$