I have read the following statement:

"The Einstein-Hilbert action can be seen as an effective field theory: setting $g_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}$, one gets a free Lagrangian for the field $h_{\mu\nu}$ plus interactions."

Now, how is it possible to define all the interaction terms?

EDIT: I ask apologize if my statement was appeared unclear: I have created a similar question which is more precise; please don't continue to consider seriously this one.

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    $\begingroup$ But the "procedure" does not only give the free Lagrangian; it does also produce interactions. I don't understand the question, sorry $\endgroup$ – AccidentalFourierTransform Dec 28 '16 at 14:04
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    $\begingroup$ This is sometimes called "Pauli-Fierz theory". Seconding AFT's comment above, please substantiate the claim that you only get a free Lagrangian in $h$ when writing the E-H action in terms of it. $\endgroup$ – ACuriousMind Dec 28 '16 at 14:10

The Einstein-Hilbert action for an $n$-dimensional space-time, $\mathcal M$, is given by,

$$S = \frac{2}{\kappa^2}\int_{\mathcal M} d^n x \, \sqrt{|g|} \, \mathcal R$$

where $\mathcal R$ is the Ricci scalar associated to the metric, $g_{\mu\nu}$ and $\kappa = \sqrt{32 G_N}$. We can linearise the theory, as you have noted, via $g_{\mu\nu} \to \eta_{\mu\nu} + \kappa h_{\mu\nu}$. Since $\mathcal R$ also depends upon the inverse metric, and the inverse in terms of $h_{\mu\nu}$ is an infinite series, the expansion of $S$ consists of an infinite number of terms, including self-interactions of the gauge field:

$$\mathcal L = \mathcal L^0 + \kappa \mathcal L^1 + \kappa^2 \mathcal L^2 + \dots$$

where the first few terms are,

$$\mathcal L^0 = -\frac14 \partial_\mu h \partial^\mu h + \frac12 \partial_\mu h^{\sigma\nu} \partial^\mu h_{\sigma\nu},$$ $$\mathcal L^1 = \frac12 h^{\alpha}_\beta \partial^\mu h^\beta_\alpha \partial_\mu h - \frac12 h^\alpha_\beta \partial_\alpha h^\mu_\nu \partial^\beta h^\nu_\mu - h^\alpha_\beta \partial_\mu h^\nu_\alpha \partial^\mu h^\beta_\nu + \frac14 h \partial^\beta h^\mu_\nu \partial_\beta h^\nu_\mu + h^\beta_\mu \partial_\nu h^\alpha_\beta \partial^\mu h^\nu_\alpha - \frac18 h \partial^\nu h \partial_\nu h$$

where $h := h^\mu_\mu.$ As you can see, we acquire terms of the form, $\mathcal L \sim h^m (\partial h)^n$ which correspond, in terms of Feynman diagrams, to self-interactions of the graviton. We can also identify a kinetic term for the theory, to extract the propagator. This theory can be coupled to others, such as a scalar, fermion, massive vector boson and the electromagnetic field.

In all such cases it is a non-renormalisable theory, but nevertheless as you have stated can be viewed as an effective field theory.

Take the Lagrangian to zeroth order in $h$ for simplicity, that is,

$$\mathcal L^0 = -\frac14 \partial_\mu h \partial^\mu h + \frac12 \partial_\mu h^{\sigma\nu} \partial^\mu h_{\sigma\nu}.$$

Under a diffeomorphism, $x^\mu \to \xi^\mu(x)$, infinitesimally the metric changes as $h_{\mu\nu} \to h_{\mu\nu} + \mathcal L_\xi h_{\mu\nu}$ which is explicitly, $\delta h_{\mu\nu} = \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu$. I now encourage you to apply this to the Lagrangian to check its behaviour under diffeomorphisms.

  • $\begingroup$ First of all thanks for this complete answer: I have checked that $L^{0}$ is gauge invariant as you suggest, but I don' t know how it is possible to extend this argument to the entire series: It seems to me the procedure violates the basic notion of GR that metric is dynamic and there is no favoured background like $\eta_{\mu\nu}$, am I wrong? $\endgroup$ – Yildiz Dec 28 '16 at 15:50
  • $\begingroup$ @Yildiz The metric is dynamical, we still have $g_{\mu\nu}$ present, but we have chosen to expand it around $\eta_{\mu\nu}$, so all the dynamical stuff is captured in the perturbation $h_{\mu\nu}$. A way to understand this if you've done quantum field theory would be to look at the background field method. Had we just set $g_{\mu\nu}=\eta_{\mu\nu}$, we would have made it non-dynamical, but obviously this isn't perturbation theory. $\endgroup$ – JamalS May 22 '20 at 17:27

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