I have a question about deriving the coupling term of string and the gauge field on brane. According to David Tong's lecture note p184/(191 in acrobat), the coupling is given by

$$ S_{\mathrm{end-point}}=\int_{\partial M} d \tau A_{a}(X) \frac{d X^a}{d \tau} \tag{1} $$

It is said that the coupling is obtained by exponentiating the vertex operator, "as described at the beginning of Section 7", $$ V_{\mathrm{photon}} \sim \int_{\partial M} d \tau \zeta_a \partial^{\tau} X^a e^{ i p \cdot x} \tag{2} $$

My question is about the logic of exponentiating the vertex operator.

In the beginning of section 7 of the lecture note, in order to obtain the coupling between string and the gauge fields, the Polyakov action is extended in curved space $$ S= \frac{1}{4 \pi \alpha' } \int d^2 \sigma \sqrt{g} g^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} G_{\mu\nu}(X) \tag{7.1} $$

The coupling comes from the bending of spacetime, e.g. $$G_{\mu\nu} (X) = \delta_{\mu\nu} + h_{\mu\nu} (X) $$ $$ Z= \int \mathcal{D} X \mathcal{D} g e^{-S_{\mathrm{Poly}} -V} = \int \mathcal{D} X \mathcal{D} g e^{-S_{\mathrm{Poly}} } (1-V +\frac{1}{2} V^2 + \dots ) $$
$$ V= \frac{1}{4 \pi \alpha'} \int d^2 \sigma \sqrt{g} g^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} h_{\mu\nu}(X) \tag{7.2} $$

In order to obtain Eq. (1) from (2), where is the bending of metric?


1 Answer 1


There is no 2-D metrics here, because we are working with the boundary $\partial M$

You could imagine a standard action $S_0 = \int_{\partial M} d\tau A_a \frac{d X^a}{d \tau}$, where $A_a$ is constant.

With a small perturbation, we will have : $A_a(X) = A_a + \epsilon_a(X)$, and we have an action $S = \int_{\partial M} d\tau A_a(X) \frac{d X^a}{d \tau}$

A partition function would be $Z = \int dX e^{-S_0 - V} = \int dX e^{-S_0}(1-V +\frac{1}{2} V^2 + \dots )$, with $V = \int_{\partial M} \epsilon_a(X)\frac{d X^a}{d \tau}$

$A_a(X)$ are coherent states of photons, as $G_{\mu\nu}(X)$ are coherent states of gravitons.

In some sense, we may consider that the photon vertex operator corresponds to a very special (non-coherent) case where $\epsilon_a(X) = \zeta_a e^{ip.x}$, in the same way as the graviton vertex operator is a very special (non-coherent) case where $h_{\mu\nu}(X) = \zeta_{\mu\nu}e^{ip.x}$

  • $\begingroup$ Thanks a lot. I still have a question, how do you get $$S_0 = \int_{\delta M} d \tau A_a \frac{ d X^a}{d \tau} $$, from Eq. (3.2.3b) in Polchinski? $\endgroup$
    – user26143
    Commented Sep 12, 2013 at 15:48
  • $\begingroup$ I don't think it is possible. For an open string, we know that we have to use vertex operators on the boundary $\delta M$ (for instance, at tree level, with the disk - Polchinski 6.2.35). So $S = \int_{\delta M} d\tau A_a(X) \frac{d X^a}{d \tau}$ is the coherent version (morally exponential) of the photon vertex operator. $\endgroup$
    – Trimok
    Commented Sep 12, 2013 at 16:20
  • 1
    $\begingroup$ Minor notational comment to the answer (v1) and its comments: A boundary of a manifold $M$ is usually denoted $\partial M$ not $\delta M$. $\endgroup$
    – Qmechanic
    Commented Sep 12, 2013 at 20:35
  • $\begingroup$ @Qmechanic : Right : corrected $\endgroup$
    – Trimok
    Commented Sep 13, 2013 at 6:48

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.