5
$\begingroup$

In chapter 7.1.1. in Tong's notes about String Theory could someone sketch how can I show the statements that he makes around eq. 7.5

  • That the addition of the counterterm can be absorbed by renormalization the wavefunction and the metric

  • How does he conclude from the renormalization $$G_{\mu \nu} \rightarrow G_{\mu \nu} + \dfrac{\alpha '}{\epsilon}\mathcal{R}_{\mu\nu} $$ that the beta function equals $$\beta_{\mu\nu}(G) = \alpha ' \mathcal{R}_{\mu \nu} \quad ? $$

$\endgroup$
2
  • 1
    $\begingroup$ Look up the computation of one-loop beta functions in dimensional regularisation. $\endgroup$
    – suresh
    Commented Sep 25, 2014 at 7:23
  • $\begingroup$ Linked. $\endgroup$ Commented May 3 at 21:23

1 Answer 1

2
$\begingroup$

On account of symmetry, 2d σ-models, even though beset with an infinity of counterterms, have those regiment themselves into few geometrical tensors constrained by symmetry; and thus limit themselves to tensors, such as the Ricci, and then underlie the β-function in the conventional manner: it is not zillions of couplings that evolve, it is only the geometry, and, on highly symmetric "pion" manifolds, by just a scale or two! (For the hypersphere, below, just its radius of curvature--it puffs up to noninteracting flatness, conformal invariance, with energy. cf. Tong 7.1.2)

Both are illustrated in our paper BCZ 1985, and especially in (2.41-2.49) and in Appendix A, meant to answer just these questions. But it is a long story, to which no justice can be done in this short format.

Nevertheless, your second question has a straightforward answer, implicit in Tong's notes, and, of course, section 2 of the paper cited here.

  • In ε = d-2 dimensions, taking the "pion" fields φ to be dimensionless, but the bare metric to have dimension ε, to one loop rewrite your expression as $$ (G_{\mu\nu}/\alpha')^{(0)}=M^\epsilon ( G_{\mu\nu}/\alpha' -\frac{1}{\epsilon}R^{(1)}_{\mu\nu} ). $$ But the bare α'-full metric must be independent of the RG scale M; so operating on this equation by $M \frac{d}{dM}$ at the pole $\epsilon \to 0$ nets $$ 0= M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} - R^{(1)}_{\mu\nu}, $$ where the superscript (1) indicates the residue at the pole, and $$ M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} = R^{(1)}_{\mu\nu}. $$

Thus, in our scaled conventions, in a hypersphere ($R_{\mu\nu}=2 G_{\mu\nu}$), whose inverse radius-squared, α', decreases with scale M , asymptotic freedom manifests itself: $d\alpha'/d\ln M =-2\alpha' ^2$. Asymptotically, the sphere flattens to a conformally invariant plane.

$\endgroup$
6
  • $\begingroup$ continuing on the expression I see using the quotient rule for derivatives and your expression for the beta function of $\alpha'$ (not sure if I'm using the right terminology) I find: $$M\frac{d}{dM}\frac{G_{\mu\nu}}{\alpha'} = \frac{1}{\alpha'}M\frac{dG_{\mu\nu}}{dM}+2G_{\mu\nu} = \frac{1}{\alpha'}\beta(G)+2G_{\mu\nu}$$ Which then gives $$\beta(G) = \alpha' R_{\mu\nu}^{(1)} + \alpha' 2 G_{\mu\nu} =? \alpha' R_{\mu\nu}$$ It looks like we got something different from what we are supposed to get, or am I missing something? $\endgroup$ Commented May 3 at 20:38
  • $\begingroup$ Could you, in particular explain where the extra term, and the superscript (1) go? $\endgroup$ Commented May 3 at 20:56
  • $\begingroup$ There is no extra term. They are explained in the linked paper, (2.44) et seq, and in any creditable QFT text on dimensional regularization, such as Schwartz's , etc. You are asking question on the iterative framework of the RG... Maybe a separate question? $\endgroup$ Commented May 3 at 21:11
  • $\begingroup$ Required reading: e.g. 26.6.1 of Schwartz. The effective coupling is G/α' ... $\endgroup$ Commented May 3 at 21:43
  • $\begingroup$ with extra term I mean $\alpha' 2 G_{\mu\nu}$, I got it based on the calculation in my comment above, but it does not appear in the anwer that Tong is giving. $\endgroup$ Commented May 4 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.