The Einstein Field Equations emerge when applying the principle of least action to the Einstein-Hilbert action, and from what I understand the path integral formulation generalizes the principle of least action. What happens when you apply the path integral instead of the action principle to the Einstein-Hilbert action?


3 Answers 3


Your questions essentially amounts to ask

How do we quantize GR?

which is the starting point of quantum gravity (QG). GR is a non-renormalizable theory, at least from the traditional perspective of perturbation theory in QFT. So the path integral with the (exponentiated) Einstein-Hilbert action as weight factor cannot easily be used to make meaningful physical predictions. New approaches to QG are needed, such as e.g. string theory (ST).

  • $\begingroup$ So, the integral over the paths diverges if you do try to take it? $\endgroup$ Oct 9, 2013 at 13:52
  • $\begingroup$ @user1825464 Well, the Euclidean version of the Einstein-Hilbert action is unbounded from below, so the path integral blows up when you try it. $\endgroup$ Oct 9, 2013 at 15:29
  • $\begingroup$ Is that specific to the Euclidean version? $\endgroup$ Oct 9, 2013 at 23:25
  • $\begingroup$ @user1825464 "This unboundedness [of the Einstein-Hilbert action] is caused by the conformal mode of the metric, whose kinetic term enters the kinetic term of (both the Lorentzian and Euclidean action) with the “wrong” sign." Page 12 of arXiv:1203.3591. $\endgroup$ Sep 8, 2015 at 19:48

The other answer and its comments are on point. I just wanted to add a little something. "What happens" is a kind of vague question. Like, what would you say "happens" when we put the action for a particle into the path integral? In any case, I thought the question might partially be asking "what would it mean to do this?" In the case of QM, we integrate over the space of all possible paths for the particle; in QFT, we integrate over the space of all possible field configurations. And quantum behavior arises from the fact that paths other than the classical one contribute something to the path integral - they happen in some sense, too. Thinking along these lines, what "happens" for the EH action in the path integral is that we now integrate over the space of all metrics - the "right" one (the one that solves the EFE) is no longer the only one that contributes to the path integral.

  • $\begingroup$ I think "What happens" is a perfectly unambiguous question. The asker clearly understand what the path integral means; (s)he is asking what kind of results we get if we do that. $\endgroup$
    – Javier
    Sep 8, 2015 at 17:33
  • $\begingroup$ Not interested in arguing the point but I'm not sure what makes it clear to you that the asker understands what the path integral means. All they say is "from what I understand the path integral formulation generalizes the principle of least action." That's hardly exhaustive. They very well might know all about it, but it wasn't clear to me from the post. $\endgroup$
    – gn0m0n
    Sep 8, 2015 at 20:57
  • $\begingroup$ Besides, what's the harm in the answer even if they did know this? It might be of interest to someone else, and if not, please, by all means ignore it :) $\endgroup$
    – gn0m0n
    Sep 8, 2015 at 20:59
  • $\begingroup$ Don't get me wrong - I wasn't chastising them, and there's nothing wrong with not being familiar with some concept. "[F]rom what I understand" just doesn't sound like someone who's formally studied it. And then there's the question in the comments about the Euclidean action, which suggests they may not be familiar with Wick rotation (and how unboundedness from below in the Euclidean version has consequences in the Minkowski version). But really, we're just speculating with limited evidence about what someone knew or didn't know. I think I've spent enough time on that now. Cheers! $\endgroup$
    – gn0m0n
    Sep 8, 2015 at 21:23

There are many issues associated with the path integral definition of the gravitational action, but here is one in particular :

Path integrals tend to be rather ill defined in the Lorentzian regime for the most part, that is, of the form

\begin{equation} \int \mathcal{D}\phi(x) F[\phi(x)]e^{iS[\phi(x)]} \end{equation}

Due to the fact that the integral oscillates, being a complex phase. To make them converge, a real factor is introduced, either by slight rotation of time ($t \rightarrow t(1 + i\varepsilon)$), or going all the way to Euclidian spacetime ($t \rightarrow it$). This gives the Euclidian path integral

\begin{equation} \int \mathcal{D}\phi(x) F[\phi(x)]e^{-S_E[\phi(x)]} \end{equation}

To converge properly, it is usually required that $S_E$ be positive. This is not the case for the gravitational action, which, in Euclidian form, is

\begin{equation} S_E = -\frac{1}{16\pi G} \int d^nx \sqrt{g} R_E(x) \end{equation}

It can even be made arbitrarily negative since a conformal transformation involves a term of the form

\begin{equation} -\frac{6}{16\pi G} \int d^nx \sqrt{g} \Omega_{,a}(x) \Omega^{,a}(x) \end{equation}

Since the conformal factor is more or less arbitrary, and you have to integrate over those configurations, it is a rather big problem to show convergence of the action.

  • 2
    $\begingroup$ $S_E$ being positive does not necessarily imply that the path integral converges. $\endgroup$
    – gented
    Oct 7, 2015 at 10:10
  • $\begingroup$ No but it is most certainly helpful. $\endgroup$
    – Slereah
    Oct 7, 2015 at 10:13
  • $\begingroup$ Not really, because what matters for renormalisation are the physical dimensions of the coupling constant. If you invert the signs making it positive it will still not converge for the same reasons as if it were negative. $\endgroup$
    – gented
    Oct 7, 2015 at 10:15
  • $\begingroup$ While not the only problem in path integrals of the gravitational action it is one to consider in such things : sciencedirect.com/science/article/pii/055032137890161X $\endgroup$
    – Slereah
    Oct 7, 2015 at 10:54
  • $\begingroup$ Probably quite stupid of me, but how did you find the equation for the euclidean action? I would just get one $i$ from $dt = i d\tau$ which together with the $i$ in front of $S$ gives me $-S_E$. I then still have the same sign inside $S_E$. Note that (for the examples I tried at least, namely various BH solutions) changing the sign of the time component in the metric does not affect R). Thus, I find$R_E=R$ (at least for those cases ). Where does the extra minus sign in $S_E$ come from? $\endgroup$
    – Kvothe
    Jan 21, 2018 at 12:47

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