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Without gravity we can easily switch between terms in a Lagrangian, such as $\partial\phi\partial\bar{\phi}$ and $\phi\Box\bar{\phi}$, since total derivative vanishes. But in GR we have additional $e\equiv\sqrt{-g}$ factor, for which ordinary derivative does not vanish $\partial e\neq 0$. Is it correct that in this case we will introduce additional term $\phi\partial\bar{\phi}\partial e$, when switching between $\partial\phi\partial\bar{\phi}$ and $\phi\Box\bar{\phi}$? And which of these two goes into the Lagrangian?

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2 Answers 2

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The covariant divergence of a vector is

$$\nabla_\mu V^\mu = \frac{\partial_\mu (V^\mu \sqrt{-g})}{\sqrt{-g}}$$

Meaning that adding a covariant divergence to the Lagrangian will result in the following change :

$$\Delta S = \int d^4x \sqrt{-g} \nabla_\mu V^\mu = \int d^4x \partial_\mu (V^\mu \sqrt{-g})$$

which is once again easy to see that it vanishes using integration by part. As with most other things in general relativity, the substitution $\partial \rightarrow \nabla$ makes this still work fine.

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  • $\begingroup$ I see what I missed now. I was only considering derivatives of scalars for which $\partial\equiv\nabla$. So in my example the vector is $\phi\partial\bar{\phi}$. Thanks, that helped. $\endgroup$
    – Kosm
    May 11, 2016 at 12:18
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OP is observing that in Minkowski space $g_{\mu\nu}=\eta_{\mu\nu}$, it doesn't matter whether we write $${\cal L} ~=~\sqrt{|g|}\partial\phi\partial\bar{\phi} \tag{1} $$ or $${\cal L}~=~-\sqrt{|g|}\phi\Box\bar{\phi}\tag{2} $$ for the Lagrangian density, if we don't care about total divergence terms. OP is pondering what happens in curved spacetime $(M,g)$? Actually both eqs. (1) and (2) still apply if we interpret the box $\Box$ as the Laplace-Beltrami operator of $(M,g)$. Of course, any other interpretation would not be geometrically sound.

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  • $\begingroup$ My mistake was to replace $\nabla$ with $\partial$, assuming I'm working with scalars, when in fact $\phi\partial\bar{\phi}$ is a vector now, so I must use covariant derivative. Silly mistake really. I see now these two terms are still the same in GR. $\endgroup$
    – Kosm
    May 11, 2016 at 12:27

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