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  World-sheet energy-momentum tensor and OPE

+ 3 like - 0 dislike

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 ]

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