Why doesn't renormalization with a Planck-scale cutoff work for quantum gravity?

Question

It's well known that general relativity and quantum theory break down at the Planck distance. What I don't understand is why you can't use the Planck distance as a sort of cutoff.

As I understand it the infinities you get from just plugging in the various field equations is sort of like the infinities you get from using the quantum field equations with the electron that was discovered back in the 1930s. If you calculate all the various virtual particles of the electrons they are infinite but what happened is with renormalization you could completely cancel the infinities. So that's the electron. But from what I understand is if you try to renormalize gravity and quantum field theory this sort of renormalization doesn't work for gravity even if you use a cutoff at the Planck level. Why is this?

Answer 1

Gravity is in fact an effective quantum field theory with the energy cut-off being the Planck scale $M_{Pl}$. The Einstein-Hilbert action is just the lowest order in an expansion in inverse powers of $M_{Pl}$. Higher order terms such as $R^2$ or $R_{\mu\nu}R^{\mu\nu}$ are supressed at low energies $E$ by powers of $E/M_{Pl}$. They only become important near the Planck scale (if no new physics arises in between).

There's no problem with this at low energies. As the energy increases, new terms may appear in the action and we should do new experimental measurements to get the new parameters, as in any non-renormalizable effective field theory. Of course, around the Planck scale, the power expansion breaks and we need a new description.

Answer 2

The reason that your prescription doesn't work is that the attempt to quantize classical gravity doesn'the yield a renormalizable QFT. In order for a QFT to be renormalizable there has to be a way of absorbing all of the infinite terms into one of the QFT parameters. For example in QED all of the infinite terms can be grouped either into the coupling parameter (e) or the electron mass parameter (m). Many potential QFT simply don't have this property.

Answer 3

The Schwarzschild radius of a black hole horizon is $r~=~2GM/c^2$. There is also DeBroglie's rule for wavelength of a particle with momentum $\vec p~=~m\vec v$ is $\lambda~=~h/|\vec p|$. The question one can ask is what happens if a black hole has a radius so that its circumference $c~=~2\pi r$ is equal to the quantum wavelength. The relativistic four-momentum is $$ P^\mu~=~\gamma(mc,~-m\vec v), $$ so the momentum of a particle sitting in a lab frame is $mc$. We then think of this mass $m$ as the mass of the black hole in this the radius of the event horizon. This then gives us $r~=~2Gh/\lambda c^3$. The wavelength is determined by a standing wave on the circumference of the black hole so that $$ r^2~=~\frac{Gh}{\pi c^3}~=~\frac{2G\hbar}{c^3}. $$ This is close, and the $2$ is removed by recognizing that the wavelength is doubled for a half standing wave. This gives then $r_{pl}~=~\sqrt\frac{G\hbar}{c^3}$.

What this really means is that a quantum bit or qubit can't be isolated on a scale smaller than this. There has been lots of silliness about discretizing spacetime into "grains" or cells (voxels etc) because of this. The problem is that these ideas violate Lorentz symmetry.