Suppose we have some field theory on a curved background, and the metric tensor $g_{\mu \nu} (x)$ is a smooth function of the position. For simplicity, let's consider a scalar theory with Lagrangian: $$ \mathcal{L} = -\frac{1}{2} g^{\mu \nu} \partial_\mu \phi \ \partial_\nu \phi + V(\phi) $$ In general, the Green function for this operator may look inattractive, and the expressions for loop integrals are unlikely to be treated analytically.

However, renormalisation is a $UV$-effect, and looking at the physical processes at distances, much smaller that the characteristic scale, on which $g_{\mu \nu} (x)$ changes, it will look approximately constant.

Does it make sense to apply a renormalisation procedure locally, namely:

  1. At each point $x$ - set $g_{\mu \nu}$ to be a constant
  2. When integrating by parts to get a propagator neglect all terms with derivatives acting on $g_{\mu \nu}$
  3. Diagonalize the resulting matrix (Green function) in momentum space, which would now have the form $A^{\mu \nu} (x) k_\mu k_\nu$ (no summation over $\mu, \nu$ is assumed)
  4. Apply the Feynman rules in that basic locally

As a result, I expect to have coupling constants to depend on the position $x$ in a certain way. Or one has to work with the exact Green function to obtain something meaningful?

  • $\begingroup$ Maybe related: The c and a-theorems and the Local Renormalisation Group: "The essential idea underlying the local RGE is to promote the elementary couplings of the QFT to position-dependent functions... so that they act as sources for the operators ... appearing in the Lagrangian. The RG flow of these local couplings, generated by the corresponding beta functions..., is then related to local Weyl rescalings of the metric..." (text above equation (1.4)). $\endgroup$ Aug 24, 2020 at 22:19
  • $\begingroup$ @ChiralAnomaly thanks for the reference! $\endgroup$ Aug 25, 2020 at 7:10
  • $\begingroup$ One doesn't need the exact Green's function, but one has to be clear: why/when do we expect a Green's function computed using a slowly-varying $g_{\mu\nu}$ to give accurate results to 1-loop in the perturbation? $\endgroup$
    – Dwagg
    Aug 26, 2020 at 2:24

1 Answer 1


Regarding your 4 point procedure: The utility of the momentum-space Feynman rules comes from translation invariance of the action, which is lost in an action with a static metric $g_{\mu\nu}(x)$ (not to mention the overall factor $\sqrt{-g}$). For instance, we don't have any momentum conserving delta functions. And neglecting all terms with derivatives acting on $g_{\mu\nu}$ while computing perturbative corrections to the Green's function seems like an uncontrolled approximation.

However, renormalization is a UV effect and something from the flat-space procedure should survive, as you mentioned. I can't give a complete answer, but I see two possible ways to proceed:

  • Standard QFT on a curved background (c.f. Carroll for instance). Pick a timelike direction, solve the classical Klein-Gordon equation (for the Gaussian truncated Lagrangian) and obtain a complete set of modes $f_i(x^\mu)$ orthonormal under the K-G inner product. The index $i$ can continuous or discrete. Expand the field $\phi = \sum_i (a_i f_i + a_i^* f_i^*)$ and quantize it as usual. The Green's function is $G(x,y) = \sum_i f_i(x) f_i^*(y)$. You can now proceed to do position-space Feynman rules to account for $\sqrt{-g} V(\phi)$ corrections.
  • If $g_{\mu\nu}\approx\eta_{\mu\nu}$ then you could approximate your Lagrangian as $-\frac12 \eta^{\mu\nu} \partial_\mu \phi\partial_\nu \phi + \lambda(x) \tilde V(\phi,\partial \phi)$ where $\tilde V$ now contains pieces of the kinetic term and $V(\phi)$. It seems such theories haven't been studies much (one study). But in principle there is nothing stopping you from proceeding in position-space Feynman rules. If $|\lambda(x)|$ is bounded you could even argue that perturbation theory is valid (to whatever extent it typically is). The study I cited works out the 1-loop corrections to the $\lambda x^\kappa \phi^4$ quartic coupling perturbation, where the integrals aren't too difficult and finds an RG fixed point.

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