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