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The action of GR is proportional to the Einstein-Hilbert action $$S_1=\int \sqrt{-g}R dx^4.$$ Now, $R$, contains terms of the form $\partial^2 g$. Using integration by parts, one can write this entirely in terms of first derivatives $\partial g$. The two actions are equivalent except for boundary terms. Lets call the action with only first derivatives, $S_2$.

Now, if the action of $S_2$ had been discovered first we might consider it the action for GR.

If these actions are inequivalent due taking boundary terms into account, how do we know which is the "true" action for our Universe? This surely means there are two inequivalent theories which may disagree cosmologically.

Indeed the action $S_2$ may be more useful when considering quantum gravity.

When I read papers and they talk about the boundary terms of GR I usually skim over it as seemingly irrelevant. But I assume it is not irrelevant otherwise they wouldn't have mentioned it in the first place.

If the boundary terms are satisfied for our Universe it means that $S_1=S_2$. But if not does it mean they are different theories of gravity? And if so, how do we know which one is correct?

Qmechanic
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1 Answers1

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The action you call $S_2$ was discovered first! (Or if not first, then at around the same time. Remember we had the EFE before the EH action). This action is known by a few names, the Einstein action, Schrodinger action, Gamma-squared action, etc, but it is of the form $ \propto \Gamma \Gamma $ where the second-order terms $\propto \partial \Gamma$ have been removed. As you stated, the action leads to the same equations of motion so we cannot physically differentiate between them. The reason for $S_2$ not being used, however, is that it is no longer a scalar, which we usually require for a proper action.

Moreover, the boundary terms that vanish in going from the standard Einstein-Hilbert action to this first-order action (i.e. $S_1-S_2$) are exactly what we require to vanish when deriving the usual GR field equations anyway$^1$. Therefore it will be the case that if we can obtain equations of motion from $S_1$, they will be the same as the EoM from $S_2$.

$^1$These boundary terms required to vanish are proportional to the metric derivatives $\delta \partial g$. As you know, to have well-posed boundary conditions one actually has to add the GHY boundary term to cancel these terms. When doing this, you'll still end up with $\delta S_1=0 \implies G_{\mu \nu}$ and $\delta S_2=0 \implies G_{\mu \nu}$, so nothing changes classically. Of course, the difference may arise in semi-classical calculations due to $S_1$ needing $S_{GHY}$.

These actions and the boundary terms are looked at in Padmanabhan's GR textbook, and a few of his papers too (e.g. https://arxiv.org/abs/gr-qc/0412068). You'll notice I focussed on the classical side, as I'm not sure the answer is so clear in quantum gravity. In fact, one of the Padmanabhan papers does argue that it's this $S_2$ not $S_1$ that arises naturally when considering the action of the spin-2 graviton: https://arxiv.org/abs/gr-qc/0409089.

Eletie
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