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A question about the coupling between string and gauge field $A_{\mu}$

+ 5 like - 0 dislike
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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
asked Sep 11, 2013 in Theoretical Physics by user26143 (340 points) [ no revision ]

1 Answer

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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 $h_{\mu\nu}(X) = \zeta_{\mu\nu}e^{ip.x}$

This post imported from StackExchange Physics at 2014-03-07 16:48 (UCT), posted by SE-user Trimok
answered Sep 12, 2013 by Trimok (950 points) [ no revision ]
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?

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

This post imported from StackExchange Physics at 2014-03-07 16:48 (UCT), posted by SE-user Trimok
Thank you very much.

This post imported from StackExchange Physics at 2014-03-07 16:48 (UCT), posted by SE-user user26143
Minor notational comment to the answer (v1) and its comments: A boundary of a manifold $M$ is usually denoted $\partial M$ not $\delta M$.

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

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

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