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?

This post imported from StackExchange Physics at 2014-03-07 16:48 (UCT), posted by SE-user user26143