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  A question about the coupling between string and gauge field Aμ

+ 5 like - 0 dislike
2110 views

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

Sendpoint=MdτAa(X)dXadτ

It is said that the coupling is obtained by exponentiating the vertex operator, "as described at the beginning of Section 7", VphotonMdτζaτXaeipx

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=14παd2σggαβαXμβXνGμν(X)

The coupling comes from the bending of spacetime, e.g. Gμν(X)=δμν+hμν(X)

Z=DXDgeSPolyV=DXDgeSPoly(1V+12V2+)

V=14παd2σggαβαXμβXνhμν(X)

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 (405 points) [ no revision ]

1 Answer

+ 5 like - 0 dislike

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

You could imagine a standard action S0=MdτAadXadτ, where Aa is constant.

With a small perturbation, we will have : Aa(X)=Aa+ϵa(X), and we have an action S=MdτAa(X)dXadτ

A partition function would be Z=dXeS0V=dXeS0(1V+12V2+), with V=Mϵa(X)dXadτ

Aa(X) are coherent states of photons, as Gμν(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 ϵa(X)=ζaeip.x, in the same way as the graviton vertex operator is a very special (non-coherent) case where hμν(X)=ζμνeip.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 (955 points) [ no revision ]
Thanks a lot. I still have a question, how do you get S0=δMdτAadXadτ
, 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 δM (for instance, at tree level, with the disk - Polchinski 6.2.35). So S=δMdτAa(X)dXadτ 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 M not δM.

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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