# World-sheet energy-momentum tensor and OPE

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?

## 2 Answers

1) First, looking at $(2.3.4)$, you see that $j^a$ is the coefficient of $\partial_a\rho$. An application of this $(2.3.12), (2.3.13)$. To make connection with this formalism, it is preferable to choose the variations :

$X^\mu\rightarrow X^\mu-\epsilon \rho(\sigma) v^c\partial_c X^\mu$

From this, we get :

$\partial_a X^\mu\rightarrow \partial_aX^\mu-\epsilon \partial_a(\rho(\sigma) v^c\partial_c X^\mu)$

Starting with an euclidean metric $\delta_{ab}$ (so the determinant is $1$), we have : $S = \dfrac{1}{4 \pi \alpha'} \int d^2\sigma \,(\partial^aX^\mu)(\partial_aX_\mu)$

The variation is then :

$\delta S =-\dfrac{\epsilon}{2 \pi \alpha'}\int d^2\sigma \,(\partial^aX^\mu) \partial_a(\rho(\sigma) v^c\partial_c X_\mu) \\ =-\dfrac{\epsilon}{2 \pi \alpha'}\int d^2\sigma[(v^c\partial^aX^\mu \partial_c X_\mu) \partial_a(\rho(\sigma)) + \rho(\sigma) (\partial^dX^\mu \partial_d(v^c\partial_c X_\mu))] \\=-\dfrac{\epsilon}{2 \pi \alpha'}\int d^2\sigma[(v^c\partial^aX^\mu \partial_c X_\mu) \partial_a(\rho(\sigma)) + \rho(\sigma) ( \partial_c (\frac{1}{2}v^c\partial^d X^\mu \partial_d X_\mu)]$

Now, we perform an integration by parts on the second term (where we eliminate the total derivative which gives a surface integral), so finally we have :

$\delta S= \dfrac{\epsilon}{2 \pi \alpha'}\int d^2\sigma[v^c\partial^aX^\mu \partial_c X_\mu) \partial_a(\rho(\sigma)) - \partial_c(\rho(\sigma))(\frac{1}{2}v^c\partial^d X^\mu \partial_d X_\mu)] \\ =\dfrac{\epsilon}{2 \pi \alpha'}\int d^2\sigma [v^c(\partial^aX^\mu \partial_c X_\mu -\frac{1}{2}\delta^a_c \partial_d X^\mu \partial^d X_\mu))\partial_a(\rho(\sigma))] \\ = -\dfrac{\epsilon}{2 \pi}\int d^2\sigma \, (v^c\,T_c^a)\partial_a(\rho(\sigma))$

Comparing to $(2.3.4)$ gives : $j^a = i v^c\,T_c^a$, or $j_a= i v^cT_{ac}$

2) @Prahar has already indicated you how come the expression of $T_{ab}$ from the action, just remember that : $\dfrac{ \partial \sqrt{\gamma}}{\partial \gamma^{ab}} = - \frac{1}{2} \sqrt{\gamma} \gamma_{ab}$

3) For you last qestion, remember that, in a general metric $g$, we have : $T_{ab}=-\frac1{\alpha'}:(\partial_aX^\mu\partial_bX_\mu-\frac12 g_{ab}\partial_c X^\mu\partial^cX_\mu):$

Now, look at $(2.1.6$, and you see that the choosen metrics for $(a,b)= (z, \bar z)$ has $g_{zz}=g_{\bar z \bar z}=0$. The stress-energy tensor has zero trace (by symmetry, $2 g^{z\bar z} T_{z\bar z}=0$), so finally $T_{z\bar z}= T_{\bar z z}=0$.

The only non-zero components are then $T(z)=T_{zz}= -\frac1{\alpha'}:\partial_z X^\mu \partial_z X_\mu:$, and $\bar T(\bar z)=T_{\bar z\bar z}= -\frac1{\alpha'}:\partial_\bar z X^\mu \partial_\bar z X_\mu:$

• Greatly appreciated!! – user46348 Sep 2 '14 at 19:10

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.