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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?

[edit] 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, Wolpertinger May 14 '17 at 13:34

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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 $$ (V^\mu)^2=g_{\mu\nu}V^\mu V^\nu. $$

  • $\begingroup$ You're welcome. I'll edit the answer to address your new question. $\endgroup$ – Salvatore Baldino May 11 '17 at 21:08
  • $\begingroup$ nevermind, I just found the answer to the follow up question myself. Thanks again $\endgroup$ – user_na May 11 '17 at 21:09

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