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

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

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

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  World-sheet energy-momentum tensor and OPE

+ 3 like - 0 dislike
2047 views

On p43 of Polchinski's book, it says that under the world-sheet translation $\sigma^a\rightarrow\sigma^a+\epsilon v^a$, $X^\mu\rightarrow X^\mu-\epsilon v^a\partial_a X^\mu$. And $$j^a=iv^b T_{ab},$$ $$T_{ab}=-\frac1{\alpha'}:(\partial_aX^\mu\partial_bX_\mu-\frac12\delta_{ab}\partial_c X^\mu\partial^cX_\mu):$$ Could someone please tell me where such expressions of $j^a$ and $T^{ab}$ come from? I assume that we start with the action $$S=\frac1{4\pi\alpha'}\int d^2\sigma\,(\partial_a X^\mu\partial_1 X_\mu+\partial_2 X^\mu\partial_2 X_\mu)=\frac1{2\pi\alpha'}\int d^2z\,\partial X^\mu\bar\partial X_\mu,$$ but I don't see how we can apply the definition of $T_{ab}$ from GR...

Further, on the next page, when expressed in complex coordinates, it says $$T(z)=-\frac1{\alpha'}:\partial X^\mu\partial X_\mu:,~~~~~\tilde T(z)=-\frac1{\alpha'}:\bar\partial X^\mu\bar\partial X_\mu:$$ How do you get here from the $T_{ab}$ above?

This post imported from StackExchange Physics at 2014-09-02 07:56 (UCT), posted by SE-user user46348
asked Sep 1, 2014 in Theoretical Physics by user46348 (20 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The action is $$ S = \frac{1}{2\pi \alpha'} \int d^2 \sigma \sqrt{\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X_\mu $$ The definition of the stress tensor from GR is $$ T_{ab} = \lambda\frac{4\pi}{\sqrt{\gamma}} \frac{ \delta S}{ \delta \gamma^{ab}} $$ Usually $\lambda = 1$, but different books use different conventions. I do not remember what convention Polchinski uses for the definition of the stress tensor. You can use the formula above to determine the stress-tensor.

The current $j^a$ can be obtained from the Noether procedure. However, it can be related to the stress-tensor as follows:

The definition of the stress tensor above implies that if we perform a metric deformation $\gamma_{ab} \to \gamma_{ab} + \delta \gamma_{ab}$, the action transforms as $$ S \to S + \frac{1}{4\pi \lambda } \int d^2 \sigma \sqrt{\gamma} T^{ab} \delta \gamma_{ab} $$ Now, under a coordinate change $\delta \sigma^a = \epsilon v^a \implies \delta \gamma_{ab} = \nabla_a (\epsilon v_b ) + \nabla_b (\epsilon v_a )$. Thus, we note that a coordinate transformation can be "converted" to a metric transformation. Under such a metric transformation, the action transforms as $$ S \to S + \frac{1}{2 \pi \lambda } \int d^2 \sigma \sqrt{\gamma} T^{ab} \left[ \nabla_a \epsilon v_b + \epsilon \nabla_a v_b \right] $$ where we have used symmetry of the stress tensor.

But, we are being naive! What about the transformation of $X^\mu$?? The action will undergo an extra transformation due to that. However, since the above transformation is a symmetry for $\epsilon$ constant, the effect of the $X^\mu$ transformation will be to simply cancel the term proportional to $\epsilon$ above. Only the term depending on the derivative of $\epsilon$ survives. Thus, under the full transformation, we must have $$ S \to S + \frac{1}{2 \pi \lambda } \int d^2 \sigma \sqrt{\gamma} T^{ab} v_b \nabla_a \epsilon $$ The current is then propertional to $$ j^a \propto T^{ab} v_b $$ The proportionality constant is again a matter of convention. I'm not sure what conventions are followed in Polchinski.

This post imported from StackExchange Physics at 2014-09-02 07:56 (UCT), posted by SE-user Prahar
answered Sep 1, 2014 by prahar21 (545 points) [ no revision ]

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:
p$\hbar\varnothing$sicsOverflow
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).
Please complete the anti-spam verification




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

Your rights
...