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I am trying to obtain the string equations of motion from the Polyakov action in the conformal gauge, i.e.: $$ S=T\int{d\tau d\sigma (\dot{x}^2-x^{'2})}\equiv\int{d\tau d\sigma \mathcal{L}} $$ where the dot means derivative with respect to $\tau$ and the prime with respect to $\sigma$. From varying this action with respect to $x^{\mu}$ I should obtain: $$ \frac{\delta S}{\delta x^{\mu}}=T\int{d\tau d\sigma (\eta^{ab}\partial_a\partial_bx_{\mu})}-T\left. \int{d\tau x^{'}_{\mu}} \right|^{\sigma=\pi}_{\sigma=0} $$ where $a,b$ run for $\tau$ and $\sigma$.

My attempt

I think that: $$ \frac{\delta S}{\delta x^{\mu}}=\int{d\tau d\sigma \left[ \frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial (\partial_{\tau} x^{\mu})}+\frac{d}{d\sigma}\frac{\partial\mathcal{L}}{\partial (\partial_{\sigma} x^{\mu})} \right] }=T\int{d\tau d\sigma \left[ \frac{d}{d\tau}(\partial_{\tau} x_{\mu})-\frac{d}{d\sigma}(\partial_{\sigma} x_{\mu}) \right] } $$ and now by means of some integration by parts I should get the result, but I am unable.

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closed as off-topic by ACuriousMind, user36790, Gert, honeste_vivere, user10851 Jun 7 '16 at 17:25

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Your derivation is close. The Polyakov action is $$ S[X,\gamma] = T\int d\tau\int d\sigma (-\gamma)^{1/2}\gamma^{ab}\partial_aX^\mu \partial_bX_\mu, $$ for $T = -(4\pi\alpha')^{-1}$. The variation with respect to the string $X_\mu$ then gives $$ \frac{\delta S}{\delta X^\mu} = T\int d\tau\int d\sigma\left[\frac{\delta}{\delta X^\mu}\left((-\gamma)^{1/2}\gamma^{ab}\partial_aX^\mu\right) \partial_bX_\mu\right] $$ $$ + T\int d\tau\int d\sigma\left[(-\gamma)^{1/2}\gamma^{ab}\partial_aX^\mu \frac{\delta}{\delta X^\mu}\partial_bX_\mu\right], $$ The first of these gives $$ \frac{\delta}{\delta X^\nu}\left((-\gamma)^{1/2}\gamma^{ab}\partial_aX^\mu\right) = (-\gamma)^{1/2}\nabla^2X_\nu $$ The second of these is $\partial_b\delta^\mu_\nu$ and evaluates the mean value theorem of calculus. The over all variation is then $$ \frac{\delta S}{\delta X^\nu} = T\int d\tau\int d\sigma(-\gamma)^{1/2}\left(\nabla^2X^\mu + \partial^a X_\mu\partial_b\delta^\mu_\nu\right) $$ $$ = T\int d\tau\int d\sigma(-\gamma)^{1/2}\nabla^2X^\mu - T\int d\tau(-\gamma)^{1/2}\partial^\sigma X_\mu\Big|_{\sigma=0}^{\sigma=\ell} $$

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  • $\begingroup$ I can't see the last step :S $\endgroup$ – Kirov Jun 3 '16 at 21:58

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