• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,075 questions , 2,226 unanswered
5,347 answers , 22,743 comments
1,470 users with positive rep
818 active unimported users
More ...

  What is a good mathematical description of the Non-renormalizability of gravity?

+ 8 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user Humble
asked Jan 26, 2011 in Theoretical Physics by Humble (65 points) [ no revision ]
retagged Apr 1, 2014
The CGHS model in 1+1D is renormalizable.

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user QGR
Now I understand the interest in extreme black holes a little better media.physics.harvard.edu/video/…

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user Humble

5 Answers

+ 6 like - 0 dislike

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.)

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user user566
answered Jan 26, 2011 by anonymous [ no revision ]
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.

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user Humble
+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user user346
answered Jan 26, 2011 by Deepak Vaid (1,985 points) [ no revision ]
+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user Lawrence B. Crowell
answered Jan 26, 2011 by Lawrence B. Crowell (590 points) [ no revision ]
there is a method behind your madness :D +1

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user user346
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 ??

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user Jose Javier Garcia
+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:30 (UCT), posted by SE-user user1355
answered Jan 26, 2011 by anonymous [ no revision ]
+ 1 like - 1 dislike

At the risk of being stoned for my reply, I will quickly tell you what follows from my experience.

Let us look at electrodynamics: when the source $J$ is a known function (operator), then the EMF is calculated without problem; no infinities, no "scales", etc., arise.

In the opposite case, when the EMF is a known field (operator), its influence on a probe charge is also calculated without difficulties.

The problem arises eprecisely when we are trying to obtain a "joint" description or to "couple" unknown charge variables with unknown field ones. Our way of coupling $J\cdot A$ brings immediately disagreement with experiment which manifests itself via unreasonably big perturbative corrections.For QED and some othe models these perturbative corrections can be represented as corrections to some coefficients of the non-perturbed (good) equations, so we discard (subtract, absorb) these divergent (or not divergent, regularized), but completely unnecessary corrections and call this "renormalization". (By the way, in CED the divergence of the self-mass is due to "self-induction" imposed by our way of coupling.)

By chance, after renormalization, we may obtain a better agreement with experiment in higher orders. P. Dirac considered this "success" a fluke, though. I wrote a toy model to demonstarate his idea..

What happens in non-renormalizable theories? Our way of coupling modifies not only coefficients in the non-perturbad equations, but also adds new equation terms, which were not present before and which spoil the solutions (agreement). The only systematic way to cope with such unnecessary and harmful terms is to keep them (artificially) small, i.e. stay away from Plank scale, for example, in our cut-off regularization.

In other words, coupling good equations together is not so trivial task, and it is namely here where we make conceptual and mathematical errors. No "scale physics" hides behind it.

answered Aug 4, 2018 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Aug 4, 2018 by Vladimir Kalitvianski

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights