I'm working through an article that has some questionable assertions. The article is by Frank Tipler, "The structure of the world from pure numbers". (I'm going to ignore the fact that some of Tipler's ideas are appropriately, I think, labeled pseudoscience.) There's a (questionable, I think) claim that "requiring the joint mathematical consistency of the Standard Model of particle physics and the DeWitt–Feynman–Weinberg theory of quantum gravity can resolve the horizon, flatness and isotropy problems of cosmology." I'm not worried about that right now. (But I imagine that won't prevent folks from commenting.) My goal is less ambitious for now. I'll just analyze, with your help hopefully, whether the following claim is correct.

In section 5 of the Rept. Prog. Phys. article cited above, he claims (quoting verbatim):

So basic quantum field theory quickly forces upon us the general invariant action \begin{align} S &= \int d^4\!x\sqrt{-g}\Big[\Lambda+\frac{1}{8\pi G}R+c_1^2R^2+c_1^3R^3+\cdots+ \nonumber \\ &+c_2^2R_{\mu\nu}R^{\mu\nu}+\cdots +c_1^3R_{\mu\nu;\alpha}R^{\mu\nu;\alpha}+\cdots \end{align}

This is the qualitatively unique gravitational Lagrangian picked out by quantum mechanics.

Physicists do not like it because (1) it has an infinite number of (renormalizable) constants $c_j$ , all of which must be determined by experiment and (2) it will it not yield second order differential equations which all physicists know and love. But the countable number of constants are in effect axioms of the theory, and I pointed out in an earlier section that the Lowenhein–Skolum theorem suggests there is no real difference between a theory with a countable number of axioms and a theory with a finite number of axioms. The finite case is just easier for humans to deal with, provided the ‘finite’ number is a small number. Further, as Weinberg (1995, pp 499, 518–519) has emphasized, this Lagranian generates a quantum theory of gravity that is just as renormalizable as QED and the SM.

The first thing I note is that, I think, the renormalization constants (we call them "low-energy constants" in hadron theory) should be different for each term. For example, as $\cdots + c_1R^2 + c_2 R^3 + \cdots$, etc.

The next thing I note and would like to focus on for this question is the claim that a countably infinite number renormalization constants, the $c_j$, isn't a problem. (I don't know what the Lowenhein–Skolum theorem is but I don't think it's crucial to this discussion.) The counterargument is that an infinite number of constants would require progressively more experimental observables to fix them as the energy scale of the probe increases. I believe that what saves the day in hadronic effective field theories is that there is a power-counting that orders the sizes of the terms in the Lagrangian density. And that this power-counting requires the determination of a finite number of low-energy constants.

Now the curvature scalar, $R = R_\mu^\mu$, is a sum of terms, the highest derivative order of which are second order in the derivatives of the metric, $g_{\mu\nu}$. And powers of $R$, say $R^n$ will have products of highest order that are $n$ factors of second derivatives of the metric. And the other terms, like $R_{\mu\nu}R^{\mu\nu}$ just have higher order derivatives. Presumably, the same power-counting logic applied above can be applied to these terms.

I believe this argument indicates that the quantum theory of gravity is predictive when viewed as an effective field theory. So my questions are: 1) Have I made the correct argument regarding the power-counting of the theory? 2) Is this argument predictive?

  • $\begingroup$ What theory is not renormalizable in the sense that you can define it using an infinite number of couplings? $\endgroup$ Dec 19, 2012 at 18:51
  • $\begingroup$ A no-go theorem for higher order derivatives theories: the energy for higher-order derivative theories become unbounded from below. For a great review of this theorem, see section 2 of arxiv.org/abs/astro-ph/0601672 $\endgroup$ Dec 19, 2012 at 20:53
  • $\begingroup$ Thanks, Jerry. I guess you meant it rhetorically. But doesn't that perspective ignore the power-counting argument? $\endgroup$
    – MarkWayne
    Dec 19, 2012 at 23:36


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