By now everybody knows that gravity is non-renormalizable, what is often lacking is a simplified mathematical description of what that means. Can anybody provide such a description?


How about a description in words instead? I think the mathematics is not the point so much. Other answers lay out the mechanics in detail. The short answer is that the theory has infinitely many divergences which cannot be absorbed by fixing finitely many couplings to their observed values. This means it is not predictive, or in more fancy words does not provide algorithmic compression: the amount of input it requires equals the amount of output it produces.

But the talk about divergences masks the physics of the situation, which I'll try to describe:

Any theory has couplings which describe the possible interactions of the system. When you probe the system at different length scales, or energy scales, those couplings change in a calculable way. This is the process of renormalization, which fundamentally has nothing to do with infinities.

Renormalizable theory is one that can be continued to high energies, or short distances, without encountering any difficulties. This means all couplings remain small, and all calculations are reliable. In non-renormalizable theories, the higher energies you probe the system, the stronger the couplings become, at some stage they become infinite. This means that you cannot do reliable calculations, which is normally an indication that there are some physical effects that you are missing. There are many examples where such physical effects are well-understood, for example new degrees of freedom (which are not visible at low energies) become important at those energy scales.

In gravity, when you calculate in perturbation theory, for small perturbations around flat space, it looks like the coupling becomes strong and something is missing. The mechanics of seeing this is straightforward and not that illuminating. Perhaps one intuitive way to understand this is that in gravity the interaction coupling is governed by the energy, so almost by definition it grows with energy...Most likely scenario is that we need new degrees of freedom to complete the theory at short distances. One suggestion for "UV completion" is string theory.

There is another scenario which may apply to gravity. Once the couplings become strong, you cannot any more reliably calculate anything, and that includes their scale dependence. So it is a logical possibility that they will stabilize at some finite value, and don't go all the way to infinity. This scenario is called asymptotic safety. In my mind this is very unlikely, and there is no indication* this scenario applies to gravity, but it remains a logical possibility.

(*I know about the various claims, this is a statement of my personal opinion.)

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    $\begingroup$ For future readers, be sure to look at answers provided by space_cadet and Lawrence B Crowell, I would say that its the totality of the three that answer the question. $\endgroup$ – Humble Jan 28 '11 at 11:31

One of the simplest ways I know of to see the non-renormalizability of gravity is to note that the coupling constant in this case (Newton's constant $G$) is dimension-full with dimensions of $L^2$.

The action for GR is of the form:

$$ S_{EH} = \frac{1}{8\pi G} \int d^4 x \sqrt{-g} R $$

When we try to expand this action as a power series in $1/G$ we see that each additional term has an extra power of $1/L^2 \sim k^2$, rendering the integration over momenta divergent in each term.

This is in contrast to QED where the coupling constant is the fine structure constant $\alpha = e^2/\hbar c \sim 1/137$ which is a dimensionless number.

Edit: For a more comprehensive and informed account see @Lawrence's answer below.


The number of divergences is not finite. In QED and other YM theories the divergences are finite and may be absorbed into the renormalization of mass or gauge charge. The rules for anomaly cancellations may be established in a finite description as well.

With quantum gravity in a QFT setting has a quadratic momentum dependence on the graviton vertex $V[k]~\sim~k^2$. This is due to the form the action takes in an expansion of the metric density ${\tilde g}_{\mu\nu}~=~g^{1/2}g_{\mu\nu}$ with $$ {\tilde g}_{\mu\nu}~=~\eta_{\mu\nu}~+~\kappa^2{\phi^\alpha}_\mu\phi_{\alpha\nu}. $$ The action $L~\sim~\kappa^{-2}\phi^{\alpha\mu;\alpha}{\phi_{\alpha\mu}}^{;\alpha}$ gives a three point vertex function which is quadratic in the momentum.

A general Feynman diagram will also have internal lines $I[k]~\sim~1/k^2$ and loops with $L[k]~\sim~\int^kd^4p$. So the internal portion of a Feynman diagram will have internal lines, vertices and loops. The Euler characteristic for a graph $$ 1~=~V[k]~-~I[k]~+~L[k] $$ is used in conjunction with the degree of divergence of these parts of the graph $D_V~=~2$ $D_I~=~-2$ $D_L~=~4$ with a total divergence $D~=~4L[k]~-~2I[k]~+~2V[k]$, so that $$ D~=~2(L[k]~+~1). $$ Consequently the divergence has an unbounded growth with the order of each Feynman diagram.

As pointed out above the gravitational constant has units of $[G]~=~Area$, which differs from the fine structure constant $\alpha~=~e^2/\hbar c$ and other gauge couplings which are unitless. The dimensional content of the gravitational constant is related to the problem of quadratic vertices. The interest in holography and AdS/CFT correspondence is with how gravity in an AdS space with negative Gaussian curvature may be replaced by quantum field on the boundary. So the divergence in a naïve QFT theory of gravity may be substituted with a stringy theory on the boundary of a space where these divergences do not exist.

  • $\begingroup$ there is a method behind your madness :D +1 $\endgroup$ – user346 Jan 26 '11 at 18:01
  • $\begingroup$ i have a question.. whenever you say that the nunmber of divergence are not 'finite' you say that there are many divergent integrals of the form $ \int d^{4}k. k^{n} $ with n=0,1,2,3,... and so on. when you say that the number of divergences is 'finite' you are referring to the fact that you have only 2 or 3 different types of divergent integrals i am right ?? $\endgroup$ – Jose Javier Garcia Feb 2 '12 at 20:42
  • $\begingroup$ Can you explain why G has units of area $m^2$. A quick search reveals that the units of G are $6.674×10^{−11} m^3 kg^{−1} s^{−2}$. Is G here not the gravitational constant? $\endgroup$ – Alexandre H. Tremblay Aug 10 '19 at 13:50

The simplified way to understand gravity's non-renormalizability is to realize that one needs to make an infinite number of infinite scaling which is not possible. In general, if the number of counterterms in the Lagrangian, required to cancel the divergences is infinite then the process of renormalization fails.


Often infinite perturbative corrections are a sign of a "bad start" in the perturbation theory (providing the exact equations are correct). In Physics one often starts from incorrect equations too and "repairs" the wrong solutions "on go", for example, by discarding infinite corrections to the fundamental constants. It "works" only in case of very simple "perturbation" Hamiltonians. In GR it is not the case (as in many other non-renormalizable QFT).

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    $\begingroup$ Dear downvoters, tell me where I am wrong, please. I would like to learn. $\endgroup$ – Vladimir Kalitvianski Jan 31 '11 at 16:22

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