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I start from the Nambu-Goto action:$$ S=-\frac{T_{0}}{c} \int_{\tau_{i}}^{\tau_{f}} d \tau \int_{0}^{\sigma_{1}} d \sigma \sqrt{\left(\dot{X} \cdot X^{\prime}\right)^{2}-(\dot{X})^{2}\left(X^{\prime}\right)^{2}} $$

I then get the Lagrangian density: $$ \mathcal{L}\left(\dot{X}^{\mu}, X^{\mu \prime}\right)=-\frac{T_{0}}{c} \sqrt{\left(\dot{X} \cdot X^{\prime}\right)^{2}-(\dot{X})^{2}\left(X^{\prime}\right)^{2}} $$

I calculate $\frac{\partial \mathcal{L}}{\partial \dot{X}^{\mu}}$ and $\frac{\partial \mathcal{L}}{\partial X^{\mu \prime}}$

$$ \frac{\partial \mathcal{L}}{\partial \dot{X}^{\mu}}=-\frac{T_{0}}{c} \frac{\left(\dot{X} \cdot X^{\prime}\right) X_{\mu}^{\prime}-\left(X^{\prime}\right)^{2} \dot{X}_{\mu}}{\sqrt{\left(\dot{X} \cdot X^{\prime}\right)^{2}-(\dot{X})^{2}\left(X^{\prime}\right)^{2}}} $$

$$ \frac{\partial \mathcal{L}}{\partial X^{\mu \prime}}=-\frac{T_{0}}{c} \frac{\left(\dot{X} \cdot X^{\prime}\right) \dot{X}_{\mu}-(\dot{X})^{2} X_{\mu}^{\prime}}{\sqrt{\left(\dot{X} \cdot X^{\prime}\right)^{2}-(\dot{X})^{2}\left(X^{\prime}\right)^{2}}} $$

Now we use the scalar product definition: $$ \boldsymbol{u} \cdot \boldsymbol{v} :=\boldsymbol{g}(\boldsymbol{u}, \boldsymbol{v})=g_{\mu \nu} u^{\mu} \boldsymbol{v}^{\nu} $$

So that $$ \dot{X} \cdot X^{\prime}=g_{\mu \nu}\dot{X}^{\mu} X^{\nu\prime} $$

So If I want to use the ADM metric for $g_{\mu \nu}$, which I found to be:$$ g_{\alpha \beta}=\left( \begin{array}{cc}{g_{00}} & {g_{0 j}} \\ {g_{i 0}} & {g_{i j}}\end{array}\right)=\left( \begin{array}{cc}{-N^{2}+\beta_{k} \beta^{k}} & {\beta_{j}} \\ {\beta_{i}} & {\gamma_{i j}}\end{array}\right) $$

Do I just put it in an multiply above or is there something more to it?

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  • $\begingroup$ Why do you want to bring the target space metric on ADM form? $\endgroup$ – Qmechanic Jun 5 at 14:40
  • $\begingroup$ To see how the action changes in curved space time and then derive the equations of motion and use different boundary conditions. $\endgroup$ – user220348 Jun 6 at 0:43

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