I'm trying to derive the following equation of motion: $$ \partial_\alpha(\sqrt{-h}h^{\alpha\beta}\partial_\beta X^\mu)=0 $$ from the Nambu-Goto action: $$ \mathcal{S}_{NG}=-\frac{1}{2\alpha}\int{d^2 x\sqrt{-h}} $$ where $\sqrt{-h}:= \sqrt{-det(h_{\alpha\beta})}$ and $ h_{\alpha\beta} := \partial_\alpha X^\mu \partial_\beta X_\mu$.

What I have done:

We use $$\frac{\delta S_{NG}}{\delta X}=0$$

Now to find the variation of the action we need to use$$ \delta\sqrt{-g}=-\frac{1}{2}\sqrt{-g}g_{\alpha\beta}\delta g^{\alpha\beta} $$

so $$ \delta S= -\frac{1}{2\alpha}\int{d^2 x(-\frac{1}{2}\sqrt{-h}\partial_\alpha X^\mu\partial_\beta X_\mu\delta h^{\alpha\beta})} $$

My problem is how is then $\frac{\delta S}{\delta X}$ going to give the form I mentioned in the beginning.


I'd like to thank @Darth_Bane for his/her beautiful answer. I'll type it here with some details for the interested reader who, like me, needs to go through it step by step.

First we note that the lagrangian has the form $\mathcal{L}=\sqrt{-h}$ , then the Euler-Lagrange equations of motion read $$ \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}\right)=\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}=0 $$ now we write this as$$ \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}\right)=\partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}\frac{\partial h_{\beta\gamma}}{\partial_\alpha X^\mu}\right) $$ then we note that $$ \frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}=\frac{\partial\sqrt{-h}}{\partial h_{\beta\gamma}}=-\frac{1}{2}\sqrt{-h}\frac{h_{\rho\sigma}\delta h^{\rho\sigma}}{\delta h_{\beta\gamma}}=-\frac{1}{2}\sqrt{-h}h^{\beta\gamma} $$ now we move on to the next fraction $$ \frac{\partial h_{\beta\gamma}}{\partial(\partial_\alpha X^\mu)} =\eta^{\mu\nu}\delta^{\alpha}_\beta\partial_\gamma X_\nu +\eta^{\mu\nu}\delta^\alpha_\gamma\partial_\beta X_\nu =\delta^\alpha_\beta\partial_\gamma X^\mu +\delta^\alpha_\gamma\partial_\beta X^\mu. $$ Putting all of this together gives $$ \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}\frac{\partial h_{\beta\gamma}}{\partial_\alpha X^\mu}\right)=\partial_\alpha(-\frac{1}{2}\sqrt{-h}h^{\beta\gamma}(\delta_{\alpha}^\beta\partial_\gamma X^\mu+\delta_\alpha^\gamma\partial_\beta X^\mu))=0 $$ $$ \frac{1}{2}\partial_\alpha(\sqrt{-h}h^{\alpha\gamma}\partial_\gamma X^\mu+\sqrt{-h}h^{\beta\alpha}\partial_\beta X^\mu)=0 $$ $$ \partial_\alpha(\sqrt{-h}h^{\alpha\beta}\partial_\beta X^\mu)=0 $$


1 Answer 1


Let me write the Nambu-Goto action as \begin{equation} S_{NG} \propto \int d^2 \sigma \sqrt{-h} = \int d^2 \sigma \mathcal{L} \end{equation}

The Euler-Lagrange equations read

\begin{equation} \partial_{\alpha}\frac{\partial \mathcal{L}}{\partial x^{\mu},_{\alpha}} = 0 \end{equation}

with the shorthand $\partial x^{\mu},_{\alpha} = \partial_{\alpha} x^{\mu}$.

We use the chain rule to write

\begin{equation} \frac{\partial \mathcal{L}}{\partial x^{\mu},_{\alpha}} = \frac{\partial \mathcal{L}}{\partial h_{\beta \gamma}} \frac{\partial h_{\beta \gamma}}{\partial x^{\mu},_{\alpha}} \end{equation}

Taking into consideration that

\begin{equation} \delta \sqrt{-h} = \frac{1}{2} h^{\beta \gamma} \delta h_{\beta \gamma} \end{equation}

as well as

\begin{equation} \frac{\partial h_{\beta \gamma}}{\partial x^{\mu},_{\alpha}} = \eta_{\mu \nu} \partial_{\beta} x^{\nu} \delta^{\alpha}_{\gamma} + \eta_{\mu \nu} \partial_{\gamma} x^{\nu} \delta^{\alpha}_{\beta} \end{equation}

yields the desired result.

I hope this was clear and it helped a bit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.