Consider the following Lagrangian density $$ \mathcal{L}(\Phi,\partial_\mu\Phi)=-\frac{1}{2}\partial_\mu\Phi\partial^\mu\Phi-\frac{m\Phi^2}{2}. $$

I want to calculate the equation of motion using the Euler-Lagrange equation to derive the Klein-Gordon equation. The E-L equation states $$\frac{\partial \mathcal{L}}{\partial\Phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\,(\partial_\mu\Phi)}=0$$ So, I calculated $\frac{\partial\mathcal{L}}{\partial\Phi}=-m^2\Phi$, and for the other term $$ \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}=-\frac{1}{2}\partial_\mu\bigg(\frac{\partial}{\partial(\partial_\mu\Phi)}\bigg)\underbrace{(\partial_\nu\Phi g^{\nu\alpha}\partial_\alpha\Phi)}_{\text{New free index, $\nu$}}=-\frac{1}{2}\partial_\mu g^{\nu \alpha}(\delta^{\mu}_\nu\partial_\alpha\Phi+\delta^\mu_\alpha \partial_\nu\Phi=-\frac{1}{2}\partial_\mu(\partial^\mu\Phi+\partial^\mu\Phi)=-\partial_\mu\partial^\mu\Phi. $$

My question is, is it correct that I have to introduce the free index $\nu$, or are there easier ways to do this?

  • $\begingroup$ It looks ok. Can you confirm the sign of the kinetic term? What metric are you using? $\endgroup$
    – DanielC
    Commented Feb 25, 2019 at 19:43
  • $\begingroup$ In your last equation, the last expression follows immediately from the first. $\endgroup$
    – my2cts
    Commented Feb 25, 2019 at 19:53
  • $\begingroup$ Probably slightly unconventional, but our professor uses $g=\text{diag}(-,+,+,+)$ $\endgroup$
    – James
    Commented Feb 25, 2019 at 19:54
  • $\begingroup$ @my2cts what do you mean? $\endgroup$
    – James
    Commented Feb 25, 2019 at 19:55
  • $\begingroup$ I mean that you don't need to use the metric tensor. $\endgroup$
    – my2cts
    Commented Feb 25, 2019 at 19:58

1 Answer 1


My question is, is it correct that I have to introduce the free index $\nu$ , or are there easier ways to do this?

You didn't introduce a free index $\nu$ as it is contracted with another one. A free index means that the index is not summed over. What you did was introduce a $dummy$ index to sum over.

The answer to your question, however, is no there is not an easier was to do this. In quantum field theory, this is as easy as it gets. The way to prove the derivative of $(\partial \Phi)^2$ is to do precisely what you did by using the definition $\partial_\mu \Phi \partial_\nu \Phi g^{\mu\nu}$ and then use the product rule.

The resulting Klein Gordon equation should not depend on what convention you use for the metric, as you can just multiply by a minus sign to get the relative minus signs correct.

The above, however, may give you the impression that equations of motion will not "look" different in different metrics. But that would be wrong. Maxwells equations are a fine example of this.

$$ \partial_\mu F^{\mu\nu} = j^\nu \qquad (+---) $$


$$ \partial_\mu F^{\mu\nu} = -j^\nu \qquad (-+++). $$

The equations look different but they are not. Indeed the "extra" minus sign in the latter equation comes from the fact that $A_\mu = (-\Phi, \mathbf{A})$ in the $(-+++)$ convention. So the equations are identical. I think that this answer may be of some use.


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