The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.

Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?

  • 2
    $\begingroup$ $\partial_\mu \phi \partial^\mu \phi$ always means $(\partial_{\mu}\phi)(\partial^{\mu}\phi)$. I have never seen it mean $\partial_\mu ( \phi \partial^\mu \phi )$ $\endgroup$
    – Prahar
    Commented Mar 27, 2017 at 17:57
  • $\begingroup$ You seem to think the implicit bracketing in $\partial_\mu \phi \partial^\mu \phi$ is $\partial_\mu (\phi\partial^\mu \phi)$. Why? $\endgroup$
    – ACuriousMind
    Commented Mar 27, 2017 at 17:57
  • $\begingroup$ @ACuriousMind If I had $x(t)$, $y(t)$ and $z(t)$ then given $\frac{d}{dt}xyz$ I would assume it meant $\frac{d}{dt}(xyz)$ and not $(\frac{d}{dt}x)(yz)$, so I've used that logic here... $\endgroup$
    – Watw
    Commented Mar 27, 2017 at 18:00

1 Answer 1

  1. The notations $(\partial\phi)^{2}$ and $(\partial_{\mu}\phi)^{2}$ are shorthands for $(\partial_{\mu}\phi)g^{\mu\nu}(\partial_{\nu}\phi)$, where $g^{\mu\nu}$ is the inverse metric tensor.

  2. Generally speaking, different authors use different notation concerning how far derivatives act to the right. When in doubt, it is a good idea to insert extra parentheses or spaces in the notation. See also this related Phys.SE post.

  3. However, in the specific context of OP's kinetic term $\partial_{\mu}\phi\partial^{\mu}\phi$, it seems universally accepted that it is supposed to mean $(\partial_{\mu}\phi)\partial^{\mu}\phi$ as opposed to $\partial_{\mu}(\phi\partial^{\mu}\phi)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.