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Nambu-Goto action

$$ S=-\mu\int\gamma^{1/2}d^2\xi $$ where $\gamma_{AB}=g_{\mu\nu}X^{\mu},_A X^{\nu},_B$ . Euler-Lagrange equations:

$$ \frac{\partial}{\partial\xi^A} \left(\frac{\partial\gamma^{1/2}}{\partial X^{\mu},_A}\right)-\frac{\partial\gamma^{1/2}}{\partial X^{\mu}}=0 $$

The variation of $\gamma^{1/2}$: $$ \delta\gamma^{1/2}=\frac{1}{2}\gamma^{1/2}\gamma^{AB}\delta\gamma_{AB} $$ My question is how to variate the induced metric $\gamma_{AB}$? I guess it should be variated with respect to spacetime coordinates and and spacetime metric. $$ \delta\gamma_{AB}=\delta(g_{\mu\nu})X^{\mu},_A X^{\nu},_B+2g_{\mu\nu}\delta(X^{\mu},_A) X^{\nu},_B $$ And the second question is, if the variation is true, then, when the derivative of $\gamma_{AB}$ with respect to $X^{\mu},_A$ and $X^{\mu}$ is evaluated must be $\delta g_{\mu\nu}=0$ and $\delta(X^{\mu},_A)=0$, similarly to the derivative of multivariable function?

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  • $\begingroup$ I have a problem with differentiation of $\gamma^{1/2}$ wrt. $X^{\mu}_A$ and $X^{\mu}$. $\endgroup$
    – Constantin
    Commented Jan 21, 2020 at 8:16
  • $\begingroup$ Crossposted to math.stackexchange.com/q/3516884/11127 $\endgroup$
    – Qmechanic
    Commented Jan 21, 2020 at 23:49

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OP is essentially asking how the Nambu-Goto (NG) action principle works. We should stress that there is no variation wrt. to the target space metric $g_{\mu\nu}(X)$ per se: It is treated as a background metric. The variation is purely wrt. the target space coordinates fields $X^{\mu}(\xi)$. It is straightforward exercise to derive the Euler-Lagrange (EL) equations for $X^{\mu}(\xi)$. For the Dirac-Bergmann constraint analysis of the NG action, see e.g. this Phys.SE post.

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  • $\begingroup$ I understood the Nambu-Goto action. The aim is to write the EOM in an explicit form evaluating the derivatives of Lagrangian density. $\endgroup$
    – Constantin
    Commented Jan 20, 2020 at 20:30
  • $\begingroup$ I am trying to bring to the following form to evaluate the derivatives. $\delta S(x,\dot{x})=\frac{\partial F}{\partial x}\delta x+\frac{\partial F}{\partial \dot{x}}\delta \dot{x}$ $\endgroup$
    – Constantin
    Commented Jan 20, 2020 at 20:41

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