Why is quantum gravity non-renormalizable?

Question

The book The Ideas Of Particle Physics contains a brief treatment of quantum gravity, in which the claim is asserted that if one attempts to construct a model of gravity along the same lines as QED, the result is non-renormalizable and the reason why can be traced to the fact that in this case the force-carrier (the graviton) is "charged" in the sense that it contains energy and therefore couples to energy which includes other gravitons. This is in contrast to the photons in QED, which are uncharged and therefore do not couple with each other.

Is this assertion accurate, or is it instead an oversimplification of something more complex?

Answer 1

As the other answer said, having a carrier that is charged has nothing to do with renormalizability. I would say that what the book says is not a mere oversimplification but it's straight up wrong.

The reason why gravity is non renormalizable has to do with the mass dimension of the coupling. To be more precise one should say that the theory is not renormalizable by power counting. There are two ways to approach this, which, in my opinion, are equally important.

Renormalization as a procedure to cure divergences

When computing Feynman diagrams sometimes it is possible to get an ill defined, divergent, answer. The purpose of renormalization is to figure out how to make sense of this.

The idea is that I start with a Lagrangian as a function of some couplings and modify the theory with a parameter $\Lambda$ so as to get finite results. Then I carefully tweak the parameters $g_i$ of my Lagrangian so that the dependence on $\Lambda$ cancels from all physical observables. In other words I have a machinery $\mathcal{F}_\Lambda$ (the Feynman diagrams) that from $\mathcal{L}_0(g_i)$ (the Lagrangian) produces the observables $f_j$ $$ \mathcal{L}_0(g_i)\;\to\;\boxed{\mathcal{F}_\Lambda}\;\to\;f_j(g_i,\Lambda)\,, $$ and I choose $g_i$ so that $f_j$ actually does not depend on $\Lambda$ at all. The problem is that this is not always possible and sometimes we need to introduce other couplings to the Lagrangian $$ \mathcal{L}_1(g_1,\ldots g_{n+1}) = \mathcal{L}_0(g_1,\ldots g_n) + g_{n+1}\mathcal{O}\,. $$ This new coupling wasn't there at the beginning, but it's needed to cancel the $\Lambda$ dependence. Every time I make a computation to a higher degree of accuracy, I risk of having to add more and more couplings. So is there any hope that this procedure stops at some point?

Yes, the answer is power counting. There is a nice property of the divergences encountered in Feynman diagrams: if the couplings that enter in the diagram have mass dimension $\delta_i = g_i$, then the divergent part can be absorbed by couplings with dimension greater or equal to $\sum_i \delta_i$.

Clearly $\delta_i \leq d$$\;{}^{\underline{1}}$ because there are no operators with negative mass dimension. So if all $\delta_i$'s are positive, the couplings are closed under renormalization. I can consistently eliminate all divergences by (in the worst case scenario) putting all possible operators of dimension $0 \leq \delta_i \leq d$.

If, on the other hand, at least one of the $\delta_i$ is negative, then there is a diagram that needs an operator whose coupling has dimension $2 \delta_i$. Which is even more negative, so we need another one with $3\delta_i$ and so on. In this scenario the procedure does not have an end and we and up with an infinite number of couplings $$ \mathcal{L}_1 =\mathcal{L}_0 + g_{n+1} \mathcal{O} + g_{n+2} \mathcal{O}' + \ldots\,. $$ We need infinitely many experiments $f_j$ to fix all of those $g_i$'s, so the theory is useless.

Renormalization group approach

Another complementary approach is that of the renormalization group. The renormalization group approach studies the behavior of a quantum system when we "zoom" out. That means when we ignore the microscopic details and retain only those dynamical variables that describe the physics at larger scales.

The net effect of doing these transformations is a change of the couplings in the Lagrangian and possibly the addition of new ones. Very much like we do in the renormalization process.

This procedure is one way obviously because we lose information in the process. Nevertheless one can try and think of it backwards. The operators whose couplings have $\delta_i > 0$ are eigenvectors of this transformation with eigenvalue smaller than one. So they become less and less important while going at small distances (high energies). On the other hand, operators with $\delta_i<0$ blow up in the high energy regime. So in order to trace them backwards we need to know with extremely high accuracy all couplings of all these operators.

This is another signature of the fact that theories with couplings of negative mass dimensions cannot be extrapolated to high energy without having to supply an infinite amount of information.

So what about gravity?

Yeah, as the other answer pointed out, gravity does have a coupling with negative dimension and it's the Newton constant (or equivalently the Planck mass to the $-2$)$\,{}^{\underline{2}}$ $$ 8 \pi G = M_P^{-2}\,. $$ But it's not all lost. As I tried to explain in the last paragraph, the problem of non renormalizability is actually an issue of high energies. The theory remains predictive at the energies we can reach in the collider. However, at energies larger than $M_P$ we have no clue.

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$\;\;{}^{\underline{1}}$ The number of dimensions, i.e. $4$.

$\;\;{}^{\underline{2}}$ I'm using Natural units.

Answer 2

I think this is a misleading oversimplification.

Gluons carry color charge and couple to themselves, yet QCD is renormalizable.

Similarly, W bosons carry weak isospin and couple to themselves, yet electroweak theory is renormalizable.

In general, non-abelian gauge theories are renormalizable despite the fact that their force-carriers couple to each other.

The problem with gravity is that its coupling constant $G$ is not dimensionless (in units where $\hbar$ and $c$ are 1). Consequently, any perturbation expansion in $G$ will involve higher and higher powers of the Riemann curvature tensor. Rather than there being a finite number of possible “counterterms” during renormalization, as in renormalizable theories, there are an infinite number of them.

Answer 3

I feel like MannyC's excellent answer deserves a brief postscript. Ultimately the reason that the gravitational coupling has negative mass dimension is a consequence of the fact that the high-energy spectrum of GR contains black holes. A good explanation of this can be found here. So yes, technically GR does not yield a renormalizable QFT because it requires an infinite number of counterterms. But this is just a symptom of the black holes in the theory.

Answer 4

The title question is at the center of the the scientific discourse/endeavor known as quantum gravity. In this answer, we will outline/sketch the standard argument of why perturbative quantum gravity is non-renormalizable.

1. Let's for simplicity work in units where $\hbar=1=c$. We will then classify physical quantities $Q$ according to their mass dimension $[Q]$, cf. Ref. 1.

2. Recall that the components $g_{\mu\nu}$ of the metric tensor is dimensionless, cf. e.g. this Phys.SE post.

3. Recall that Newton's gravitational constant $G$ in $d$ spacetime dimensions has mass dimension $$[G]~=~2-d,\tag{1}$$ cf. e.g. this Phys.SE post.

4. Recall that the Einstein-Hilbert (EH) action in $d$ spacetime dimensions is $$ S~=~ \frac{1}{16\pi G} \int \! d^dx \sqrt{-g} (R-2\Lambda).\tag{2} $$ One may check that the action $S$ has mass dimension $[S]=0$, as it should, because the curvature $R$ contains 2 spacetime derivatives, $[R]=2$.

5. In perturbative gravity, the metric field $$ g_{\mu\nu}~=~\eta_{\mu\nu}+\kappa h_{\mu\nu},\qquad \kappa~:=~2\sqrt{8\pi G}, \tag{3}$$ is expanded around a background $\eta_{\mu\nu}$, typically Minkowski spacetime with cosmological constant $\Lambda=0$.

6. The quantum fluctuations $h_{\mu\nu}$ have been scaled so that the quadratic term in the Lagrangian density ${\cal L}$ is canonically normalized. Suppressing various tensor structures and ignoring dimensionless constants, the Lagrangian density ${\cal L}$ can be schematically written as$^1$ $$\begin{align} {\cal L}~=~&(\partial h)^2\sum_{m\in\mathbb{N}_0}(\kappa h)^m\cr ~=~&\underbrace{(\partial h)^2}_{\text{quadratic}} + \underbrace{\kappa(\partial h)^2h}_{\text{cubic}}+ \underbrace{\kappa^2(\partial h)^2h^2}_{\text{quartic}}+\ldots,\end{align}\tag{4}$$ as seen in other SE posts.

7. In the Lagrangian density (4) the coupling constant $\kappa$ of the cubic interaction term has mass dimension $$[\kappa]~=~1-\frac{d}{2}\tag{5},$$ which is negative if $d>2$, and therefore non-renormalizable in the old Dyson sense, cf. e.g. this Phys.SE post. (Note that gravity in $d=3$ is topological.)

References:

1. M. Srednicki, QFT, 2007; Chapter 12. A prepublication draft PDF file is available here.

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$^1$ One may check that each term in eq. (4) has mass-dimension $[{\cal L}]=d$, as it should. Here $$[h_{\mu\nu}]~=~\frac{d}{2}-1\tag{6}$$ is the standard mass dimension of a perturbative bosonic field in QFT.