Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,549 answers , 19,356 comments
1,470 users with positive rep
410 active unimported users
More ...

Effective action for bosonic string theory with enhanced symmetry

+ 3 like - 0 dislike
119 views

See these lecture http://members.ift.uam-csic.es/auranga/lect7.pdf page 17.

Usually one derives the effective action from the massless states calculating amplitudes, otherwise through beta functions(worldsheet conformal invariance). One obtains a effective field theory containing a metric, a Kalb-ramond field and a dilaton. These came from the $N=1,\bar{N}=1$ sector of the mass spectrum. This is wellknown.

If the 25th dimension is compactified with $R=\alpha'^{1/2}$ then another massless 25dimensional fields emerge. This happens in the $N=0,\bar{N}=1$ sector and $N=1,\bar{N}=0$ sector. After analysing the enhanced symmetry from the 25d point of view one can deduce that these are $SU(2)\times SU(2)$ gauge bosons.

On page 17 of the lecture the autor says: it is possible to cook up a new 25d effective field theory by including by hand the new massless modes.

How this action looks like? Is there any paper or reference which explains the calculations or steps to obtain the action?


This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Anthonny

asked Nov 27, 2013 in Theoretical Physics by Anthonny (70 points) [ revision history ]
edited Apr 19, 2014 by dimension10
Could you explain the steps and calculations to obtain the effective field theory action? Please, give some reference (paper, book, webpage, blog) related to this interesting topic

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Anthonny
Im waiting for more answers, nobody else? The massless sector of open string is a gauge boson, could this help? I think that the action should be nonlinear like a Born Infield action. What do you think?

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Anthonny

1 Answer

+ 2 like - 0 dislike

The value $R=\alpha^{\prime 1/2}$ is the self-dual radius under T-duality. One may indeed extract the massless spectrum – the spectrum of all fields much lighter than $\alpha^{\prime -1/2}$.

Because the CFT has an $SU(2)\times SU(2)$ symmetry, as can be seen from the OPEs of the currents, the spacetime physics has this symmetry, too. Because one finds (spacetime) Lorentz vector states in the adjoint of $SU(2)\times SU(2)$, it is clear that this group is the gauge symmetry of the spacetime physics.

And indeed, one may verify that the tree-level scattering amplitudes for all the relevant string modes agree with the scattering amplitudes extracted for quanta of fields in the effective action that is (a bit schematically, especially when it comes to the parts unrelated to the enhanced gauge symmetry) $$ S =\int d^{25}x\,\exp(2\phi) [R + (\partial_{[\lambda} B_{\mu\nu]})^2 + (\partial_\mu \phi)^2 -\frac 14 {\rm Tr}(F_{\mu\nu}F^{\mu\nu}) ] $$ So it is a 25-dimensional action because we ignore the 1 compactified dimension whose radius is stringy (dimensional reduction). In this 25-dimensional spacetime, there is the dilaton, the metric, the B-field, and an $SU(2)\times SU(2)$ gauge field, and they have more or less the expected terms in the effective action.

See Polchinski's Volume 1 from page 242 to 250+ or so. The effective action is probably not written there explicitly. However, you may find the 26D effective action for the uncompactified bosonic string theory on the top of page 114, reduce the dimension, and add the $SU(2)\times SU(2)$ Yang-Mills field, more or less getting the exact answer. The "Cartan" $U(1)\times U(1)$ part of the Yang-Mills action comes from the Kaluza-Klein $U(1)$ symmetry of the circle and from the components of the B-field $B_{\mu,25}$. This is "enhanced" by the extra "accidentally massless" states to the non-Abelian group.

Between equations 3.7.15 and 3.7.20 or so, Polchinski takes a different but ultimately equivalent strategy to derive the spacetime action. He derives the equations of motion from the requirement of the conformal symmetry on the world sheet (vanishing beta-functions etc.) and verifies that the same equations follow as the Euler-Lagrange equations from the spacetime action he "guesses" and "refines".

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Luboš Motl
answered Nov 29, 2013 by Luboš Motl (10,178 points) [ no revision ]
I was thinking of that term $-\frac 14 {\rm Tr}(F_{\mu\nu}F^{\mu\nu})$ is the most obvious. But I think its just the linear part, as the same way a linear action for $h_{\mu\nu}$(the weak gravitational field) is the linear part of $\int R$ the complete nonlinear action. For example for the open bosonic string (not compactified) the massless mode is a gauge boson and when one obtain its effective action one doesnt obtain $\int -\frac 14 {\rm Tr}(F_{\mu\nu}F^{\mu\nu} $ but a Born Infield action instead.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Anthonny
Another thing Im not sure is if the coupling with the dilaton is necesary because they came from diferent sectors, I think this is not obvious, one might intuit the action or guess an expected action. For this reason I want a reference like a thesis or research paper, so I can be sure that the calculation was done and the answer is right. Why you are sure of your response? Have you done this calculation? or where did you see that action?

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Anthonny

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...