Musings

Since my previous post was about asymptotic safety, I don’t particularly feel drawn to write another one anytime soon.

But I will leave you to puzzle over the logic in the key sentence of the paper:

The second functional derivative of (5) results in a sum of terms containing at least one power of the Weyl tensor. Since $C$ is trace-free by construction, all its contractions with the metric vanish. This entails that there is no feedback of the Goroff-Sagnotti term on the renormalization flow of Newton’s constant and the cosmological constant. [emphasis mine]

If you replace “at least” with “exactly”, then the conclusion would follow. But you can’t, so it doesn’t.

Then we lack the intellectual tools to successfully address any of these questions and the whole discussion is a complete waste of time.

The whole construction uses the background field formalism. Computing the second variation of the $C^3$-term around a background, one finds that the Hessian contains terms proportional to $C$, $C^2$ and $C^3$ (omitting tensor structures for simplicity).

And is obviously nonzero.

The beta-function of Newtons constant is encoded in the coefficient proportional to the Ricci-scalar of the background metric. None of the terms appearing above has the structure to give such a contribution.

Umh. No. You need to stick that Hessian into (2), where it surely contributes to the $\beta$-function for Newton’s constant.

Formally this can be proven using off-diagonal heat-kernel techniques.

I would like to see that proof, as the result (that sticking the aforementioned Hessian into equation (2), of your paper, yields no contribution to the running of the Ricci scalar term in the action) surprises me greatly.

Thus the conclusion made in the key sentence is correct.

That key sentence is the crux of the whole paper, so actually presenting the argument (which you have alluded to in your comment, but not given) would be an important thing to include in the paper.

Thanks for the reference. I’ll take a look.

Let me, at least try to sketch a bit more of their argument, so that we can see whether your considerations shed some light. Following them, let’s use the electromagnetic case as a foil.

First, let’s review the Minkowski case. Let $\varepsilon(z,\overline{z})$ be an arbitrary function on $S^2$. We can extend it to a function $\varepsilon_+$ on $\mathcal{I}^+$, in a canonical way, by taking $\varepsilon_+$ to be independent of $u$ (the retarded time coordinate on $\mathcal{I}^+$). The associated charge, $$Q_+ = \int_{\mathcal{I}^+} \mathcal{J}_+$$ where $$\mathcal{J}_+ = d\varepsilon_+\wedge *F + \varepsilon_+ j$$ (and Maxwell’s equation is $d*F=j$). We can do a similar extension of $\varepsilon$ to a function on $\mathcal{I}^-$, and the soft photon theorem is the statement that the matrix elements of $Q_+-Q_-$ vanish, for any choice of $\varepsilon(z,\overline{z})$.

Now let’s turn to the case of interest, a blackhole formed by a collapsing spherical shell (denoted in the Penrose diagram, below, in red), which subsequently evaporates.

$$H_\lt$$

They want to say the following

1. There is a canonical way to extend $\varepsilon$ to a function $\hat{\varepsilon}$, defined on the Cauchy surface $R= H_\lt \cup S \cup \mathcal{I}^+_\lt$, denoted in blue in the Penrose diagram above.
2. The associated charge, $Q_R = \int_R \hat{\mathcal{J}}$, satisfies $Q_+-Q_R=0$.
3. The contribution to $Q_R$ from the integral over the spacelike surface, $S$, is somehow negligible.

One then concludes that $$\int_{H_\lt} \hat{\mathcal{J}} = \int_{\mathcal{I}^+_\gt} \mathcal{J}_+$$

I already said that I don’t see (1).

I also don’t see (3). Recall that, in Feynman diagrams, the soft theorem is obtained by attaching soft-photon (or soft-graviton) lines (both incoming and outgoing) to the external lines (in this case, the infalling matter denoted in red and the outgoing Hawking quanta) in all possible ways. “Most” of those soft lines cross the surface $S$, so I don’t see how you can ever get the correct Ward identity by neglecting the contribution from $S$.