I'm attempting to learn about string theory through a series of lectures by Freddy Cachazo. Starting 43 minutes into part 6, he begins to explain why we can associate the traceless, massless, closed string excitation with gravitation.

The Polyakov action is modified so that the metric tensor of the target space is some $G_{\mu\nu} \left(X\right)$ rather than $\eta_{\mu\nu}$, and it is shown that, in an appropriate limit, the vanishing of the world-sheet field theory's beta function (at one loop) due to conformality yields the vacuum Einstein field equations for the target space metric.

At this point, he seems to imply that the argument is finished, but I am a bit lost, as there seems to be a logical step that I am missing. Why does the emergence of the vacuum EFEs in this context tell us that this particular excitation of the string (which we found starting from the original Polyakov action) is a graviton?

  • 2
    $\begingroup$ See: physics.stackexchange.com/q/39476/50583, perhaps also physics.stackexchange.com/q/108230/50583 $\endgroup$
    – ACuriousMind
    Oct 11, 2022 at 13:01
  • $\begingroup$ Because you find that the metric changes? $\endgroup$
    – Javier
    Oct 11, 2022 at 14:14
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    $\begingroup$ Since you bountied this question, it seems you don't find the questions I linked sufficient explanation for why a spin-2 particle has to be a graviton. Can you be more specific what exactly you want to know here? $\endgroup$
    – ACuriousMind
    Oct 23, 2022 at 15:04
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    $\begingroup$ @ACuriousMind I understand that an interacting massless spin-2 particle essentially has to be a graviton, but I didn’t follow the connection presented in the lecture, i.e. replacing $\eta_{\mu\nu}$ with a general metric $G_{\mu\nu}\left(X\right)$ depending on the fields $X^\mu$ in the Polyakov action, analysing the EOMs coming from one-loop renormalisation, and concluding that the action they come from is, somehow, that of the effective field theory for the particular massless excited states originally found in the spectrum of the canonically quantised string in Minkowski spacetime. $\endgroup$ Oct 23, 2022 at 16:40

1 Answer 1


The string amplitudes for the asymptotic states are given by the Polyakov path integral with vertex operators insertions $\mathcal{V}_k$ corresponding to those asymptotic states. Consider it on the nontrivial metric background that is close to the flat spacetime, \begin{equation} G_{\mu\nu}(X)\simeq \eta_{\mu\nu}+h_{\mu\nu}, \end{equation} The path integral then may be decomposed into an infinite sum, \begin{multline} \sum_{\text{topologies}}\int \frac{DX^\mu Dg^{ab}}{Vol_{\text{Diff}\times \text{Weyl}}} \exp\Big[-\frac{1}{2\pi\alpha'}\int d^2\sigma\sqrt{g}G_{\mu\nu}(X)\partial_a X^\mu\partial_b X^\nu g^{ab}\Big]\prod_k\mathcal{V}_k\\ =\sum_{\text{topologies}}\int \frac{DX^\mu Dg^{ab}}{Vol_{\text{Diff}\times \text{Weyl}}} \exp\Big[-\frac{1}{2\pi\alpha'}\int d^2\sigma\sqrt{g}\eta_{\mu\nu}\partial_a X^\mu\partial_b X^\nu g^{ab}\Big]\prod_k\mathcal{V}_k\Big(1-\int d^2\sigma \sqrt{g} h_{\mu\nu}(X) \partial_a X^\mu\partial_b X^\nu g^{ab}\\ +\frac{1}{2}\int d^2\sigma \sqrt{g} h_{\mu\nu}(X) \partial_a X^\mu\partial_b X^\nu g^{ab}\cdot \int d^2\sigma \sqrt{g} h_{\mu\nu}(X) \partial_a X^\mu\partial_b X^\nu g^{ab}+\ldots\Big) \end{multline} This may be considered as a path integral of the string on a flat spacetime. Each term in the expansion adds the extra factor $\int d^2\sigma \sqrt{g} h_{\mu\nu}(X) \partial_a X^\mu\partial_b X^\nu g^{ab}$ which happens to coincide with the vertex operator of the massless spin 2 excitation of the string on a flat spacetime (I will call it simply graviton) I.e. the amplitude computed for the nontrivial background is equivalent to the amplitude for the asymptotic states on a trivial background PLUS the amplitude for the same asymptotic states AND one graviton PLUS the amplitude for the same asymptotic states AND two gravitons etc.

This allows us to interpret the nontrivial background $G_{\mu\nu}$ as a coherent state of the gravitons (very similar to how the nontrivial "classical" background in the field theory is related to the coherent states of the field). I.e. $G_{\mu\nu}$ is not an independent entity in the string theory but is constructed from the string excitations. The vanishing $\beta$-function provides is a self-consistency condition that this background must satisfy, otherwise you can't define properly the Polyakov path integral on such background.

This self-consistency condition happens to take the form of the classical equations of motion. In the $\alpha'\sim l_s^2\rightarrow \infty$ limit you may find the corresponding action (up to the unknown common factor that determines the Planck scale $l_P$). Then you may treat this action as the action of the effective field theory (EFT) and try to compute the amplitudes e.g. for the $n$ gravitons to $m$ gravitons scattering (that of course make sense only for $E\ll l_P$).

Now, you may compute the string amplitudes for the $n$ gravitons to $m$ gravitons scattering from the path integral above. Then take the $E\ll l_s$ limit in such amplitude. And you will see that in this limit the string amplitudes turn into the EFT amplitudes (with the unknown factor usually fixed by comparing the 3-point amplitudes). This means that indeed the self-consistency condition for the background reflects the dynamics of the string excitations the background is made of. And that the action we write for this self-consistency condition is indeed the low energy effective action for the string theory.

You may read this stuff in more detail and look for the references e.g. in Vol. 1 of Polchinski's textbook.


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