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I want to derive the conserved charge for Lorentz rotations in the target space of the following action.

$$ \mathscr{S}_p=-\frac{T}{2}\int{d^2\sigma\;\eta^{\alpha\beta}\partial_\alpha X^\mu\partial_\beta X_\mu}\tag{1} $$ the Polyakov action with fixed $h_{\alpha\beta}$.

The transformations I assume would take the form $$ \delta X^\mu=\alpha^\mu_\nu X^\nu.\tag{2} $$ I'm not quite sure how to apply $$ \mathcal{J}^\alpha=\delta\phi\frac{\delta\mathscr{S}}{\delta(\partial_\alpha\phi)}.\tag{3} $$ After searching for the form of the result it looks like it should have an anti-symmetric term and an extra term.

A detailed derivation would be really helpful!

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    $\begingroup$ Hi @fielder: Are you following a reference? Which page? $\endgroup$
    – Qmechanic
    Commented Apr 11, 2020 at 17:45
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    $\begingroup$ The $X^\mu$ are your "fields" $\phi$. What would you get? What do you mean by "looks like"? Are you asking the reader to second-guess what you fail to attribute? $\endgroup$ Commented Apr 11, 2020 at 18:16
  • $\begingroup$ @Qmechanic these relations can be found in any string theory book. I am more familiar with the BBS book and the relevant discussion is from eq.(2.64) until (2.68). $\endgroup$
    – user172341
    Commented Apr 11, 2020 at 22:43
  • $\begingroup$ @fielder, please let me know if my answer is clear or not. $\endgroup$
    – user172341
    Commented Apr 11, 2020 at 22:44
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    $\begingroup$ @Konstantinos: I know. I'm asking for a different reason. $\endgroup$
    – Qmechanic
    Commented Apr 11, 2020 at 23:05

1 Answer 1

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Recall that under the global symmetry transformation $\phi \rightarrow \phi + \delta_{\epsilon} \phi$, the Noether current can be determined via $\mathcal{L} \rightarrow \mathcal{L} + \epsilon \partial_{\alpha} \mathcal{J}^{\alpha}$.

Let me try to make this interesting and pedagogical. Let's consider the case of translations first. That is, the transformation $\delta X^{\mu} = b^{\mu}$ just to show you how it works.

The computation is presented below:

From the Lagrangian

$\begin{equation} \mathcal{L} = \eta^{ab} \eta_{\mu \nu} \partial_{a} X^{\mu} \partial_b X^{\nu} \end{equation}$

we have -if we insert the variations-

$\begin{equation} \begin{split} &\eta^{ab} ~ \eta_{\mu \nu} ~ \partial_{a} \left( X^{\mu}+b^{\mu} \right) \partial_b \left( X^{\nu} + b^{\nu} \right) = \\ &\eta^{ab} ~ \eta_{\mu \nu} ~ \left( \partial_{a} X^{\mu} \partial_{b} X^{\nu} + \partial_{a} X^{\mu} \partial_{b} b^{\nu} + \partial_{a} b^{\mu} \partial_{b} X^{\nu} + \partial_{a} b^{\mu} \partial_{b} b^{\nu} \right) \end{split} \end{equation}$

The final term is $b^2$ (in other words it is $\mathcal{O}(\delta^2)$) and we can ignore it. It is easy to see that the first term of the sum is the original Lagrangian. We can write

$\begin{equation} \mathcal{L} \rightarrow \mathcal{L} - \frac{T}{2} ~ \eta^{ab} ~ \eta^{\mu \nu} \left( \partial_{a} X^{\mu} \partial_{b} X^{\nu} + \partial_{a} b^{\mu} \partial_{b} X^{\nu} \right) \end{equation}$

Now, we can massage the terms a bit. The manipulations are fairly straightforward. We show the steps explicitly below

$\begin{equation} \begin{split} & \eta^{ab} ~ \eta_{\mu \nu} \left( \partial_{a} X^{\mu} \partial_{b} b^{\nu} \right) = \\ & \eta^{ab} ~ \eta_{\nu \mu} \left( \partial_{a} X^{\nu} \partial_{b} b^{\mu} \right) = \\ & \eta^{ab} ~ \eta_{\mu \nu} \left( \partial_{a} X^{\nu} \partial_{b} b^{\mu} \right) = \\ & \eta^{ab} ~ \left( \partial_{a} X_{\mu} \partial_{b} b^{\mu} \right) \end{split} \end{equation}$

The manipulations are: first relabel $\mu \leftrightarrow \nu$, then use the fact that the metric is symmetric and then we just lower an index.

The other term can also be massaged in the same way. We show that below:

$\begin{equation} \begin{split} &\eta^{ab} ~ \eta_{\mu \nu} (\partial_{a} b^{\mu} \partial_{b} X^{\nu}) = \\ &\eta^{ab} (\partial_{a} b^{\mu} \partial_{b} X_{\mu}) = \\ &\eta^{ba} (\partial_{b} b^{\mu} \partial_{a} X_{\mu}) = \\ &\eta^{ab} (\partial_{b} b^{\mu} \partial_{a} X_{\mu}) = \\ &\eta^{ab} (\partial_{a} X_{\mu} \partial_{b} b^{\mu}) \end{split} \end{equation}$

So, it is obvious that schematically we can write

$\begin{equation} \mathcal{L} \rightarrow \mathcal{L} + \delta \mathcal{L} \end{equation}$

where we have

$\begin{equation} \delta \mathcal{L} = - T \eta^{ab} \partial_{a} X_{\mu} \partial_{b} b^{\mu} \end{equation}$

Integrate by parts the expression and we can read off

$\begin{equation} P^{\mu}_a = T \partial_{a} X^{\mu} \end{equation}$

Let us consider the Lorentz transformations. This time we have

$\begin{equation} \delta X^{\mu} = a^{\mu}_{\nu} X^{\nu} \end{equation}$

Hopefully, the steps are clear from the previous considerations and below the computation is presented a bit more quickly. Let's do the variation of the action

$\begin{equation} \begin{split} \delta S = - T \int d^2 x \left((\partial_{a} a^{\mu}_{k}) X^{k} \partial^a X^{\nu} \eta_{\mu \nu} + a_{\nu k} \partial_{a} X^{k} \partial^{a} X^{\nu} \eta_{\mu \nu} \right) \end{split} \end{equation}$

The second term in the above vanishes due to symmetries. Therefore we have

$\begin{equation} \delta S = -T \int d^2 x (\partial_{a} a_{\nu k}) X^{k} \partial^{a} X^{\nu} \end{equation}$

from which we can obtain

$\begin{equation} \mathcal{J}^{\mu \nu}_{a} = T (X^{\mu} \partial_a X^{\nu} - X^{\nu} \partial_a X^{\mu}) \end{equation}$

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  • $\begingroup$ Thank you very much @Konstantinos, after going through it myself, this was very helpful! Could you also show how I would get the conserved charge? Or should I post a different question ? $\endgroup$ Commented Apr 12, 2020 at 15:45
  • $\begingroup$ Hi, glad I was able to help. In general, I think that here the general rule is that each post should have a self-contained and unique (one) question. I am not sure though. Maybe you should ask one of the moderators to tell you what to do, i.e Qmechanic who has also left a comment. I am swamped at the moment to provide more computations. Sorry $\endgroup$
    – user172341
    Commented Apr 12, 2020 at 16:32

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