The Polyakov action for a string in spacetime $(M, g)$ is

$$\mathcal{L} = \frac{T}{2}\sqrt{-\det(h)}h^{\eta\lambda}(\sigma)g_{\mu\nu}(x(\sigma))\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\frac{\partial x^{\nu}}{\partial \sigma^{\lambda}}$$

And Euler-Lagrange equations for a classical field reads

$$\frac{\partial}{\partial \sigma^{\rho}}\frac{\partial \mathcal{L}}{\partial(\frac{\partial x^{\alpha}}{\partial \sigma^{\rho}})}-\frac{\partial \mathcal{L}}{\partial x^{\alpha}} = 0$$

Using the fact that $\displaystyle \frac{\partial \mathcal{L}}{\partial x^{\alpha}} = \frac{T}{2}\sqrt{-\det(h)}h^{\eta\lambda}(\sigma)\frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\frac{\partial x^{\nu}}{\partial \sigma^{\lambda}}$, we conclude that

$$\frac{\partial}{\partial \sigma^{\rho}}\biggr(\frac{T}{2}\sqrt{-\det(h)}h^{\eta\rho}(\sigma)g_{\mu\alpha}(x(\sigma))\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\biggr) - \frac{T}{2}\sqrt{-\det(h)}h^{\eta\lambda}(\sigma)\frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\frac{\partial x^{\nu}}{\partial \sigma^{\lambda}} = 0$$

$$\frac{\partial}{\partial\sigma^{\rho}}\sqrt{-\det(h)} = \frac{1}{2}\sqrt{-\det(h)}h^{\gamma\beta}(\sigma)\frac{\partial h_{\gamma\beta}}{\partial \sigma^{\rho}} = \frac{1}{2}\sqrt{-\det(h)}\Gamma^{\gamma}_{\rho\gamma}$$

$$\begin{align}\biggr(-\frac{T}{4}\sqrt{-\det(h)}\Gamma^{\gamma}_{\rho \gamma}h^{\eta\rho}(\sigma)g_{\mu\alpha}(x(\sigma))\frac{\partial x^{\mu}}{\partial\sigma^{\eta}} + \frac{T}{2}\sqrt{-\det(h)}\frac{\partial h^{\eta\rho}}{\partial \sigma^{\rho}}g_{\mu\alpha}(x(\sigma))\frac{\partial x^{\mu}}{\partial \sigma^{\eta}} &+ \frac{T}{2}\sqrt{-\det(h)}h^{\eta\rho}(\sigma)g_{\mu\alpha}(x(\sigma))\frac{\partial}{\partial \sigma^{\rho}}\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\biggr) - \frac{T}{2}\sqrt{-\det(h)}h^{\eta\lambda}(\sigma)\frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}\frac{\partial x^{\mu}}{\partial \sigma^{\eta}}\frac{\partial x^{\nu}}{\partial \sigma^{\lambda}} &= 0\end{align}$$

Does this make any sense?

Note: I am looking for a derivation without using Poincaré and Weyl invariance (see section "Symmetries of the Polyakov Action" and the comments). Details of the derivation can be found here, here and here, while this is for the Nambu-Goto action.

  • $\begingroup$ You may find section 2.4.2 here useful: google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$
    – Quillo
    Jan 20 at 11:26
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    $\begingroup$ Thanks for that reference, although I had hoped for a derivation without using Poincaré and Weyl invariance to make $g$ terms disappear. $\endgroup$
    – user348860
    Jan 20 at 11:30
  • $\begingroup$ Deane, I took the freedom to link several (I hope useful) questions within your answer. If you come up with a good answer, please post your own answer. Unfortunately, I can not help you more than this because this is not really in my expertise. If I come up with a clear explicit calculation, I will post it. $\endgroup$
    – Quillo
    Jan 20 at 15:37
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    $\begingroup$ If you're allowing for an arbitrary target space metric $g(x(\sigma))$, how can you conlcude that $\frac{\partial\mathcal{L}}{\partial x^\mu} = 0$? $\endgroup$ Jan 20 at 16:31
  • $\begingroup$ 1. You have made over 30, mostly minor, edits to this question. Please be mindful that every edit bumps the question on the page of active question and try to make your edits few and substantial. 2. One of your edits removed a lot of context and links from the question. Please do not drastically change questions, especially not after they have received an answer. $\endgroup$
    – ACuriousMind
    Jan 21 at 16:12

1 Answer 1


$\newcommand{\d}{\mathrm{d}}\newcommand{\pd}{\partial}$Let's use worldsheet differential forms to make our life easier. So, for example, $$\d X^\mu(\sigma) = \pd_\alpha X^\mu(\sigma)\; \d\sigma^\alpha.$$ I will also denote the fields by capital $X$; $\left\{\mu,\nu,\ldots\right\}$ will be spacetime indices and $\{\alpha,\beta,\ldots\}$ will be worldsheet indices (whenever necessary), so expressions like $\partial_\alpha$ will mean $\frac{\partial}{\partial\sigma^\alpha}$.

With these, the Polyakov action for the fields $$X^\mu\in\operatorname{Maps}((\Sigma,h)\to(M,g))\cong \Omega^0(\Sigma;M)$$ can be rewritten as $$S[X] = \frac{T}{2}\int_\Sigma g_{\mu\nu}(X)\ \d X^\mu\wedge \star\, \d X^\nu,$$ where $\star$ is the Hodge star operator on $\Sigma$. Note that from the point of view of the worldsheet, $\Sigma$, $g_{\mu\nu}$ is a scalar. It is then easy to just vary the action and get \begin{align} 0\overset{!}{=} \delta S &= \frac{T}{2}\int_\Sigma \Big(\delta g_{\mu\nu}(X)\ \d X^\mu\wedge \star\, \d X^\nu + 2\,g_{\mu\nu}(X)\ \d\ \delta X^\mu\wedge \star\, \d X^\nu \Big) = \\ &= \frac{T}{2}\int_\Sigma \Big(2\, \Gamma^{(g)}_{\mu\nu\lambda}\ \d X^\mu\wedge \star\, \d X^\nu - 2\, \d\big(g_{\mu\lambda}(X)\ \star\, \d X^\mu\big) \Big)\ \delta X^\lambda, \end{align} where $$\Gamma^{(g)}_{\mu\nu\lambda}:=\frac12\left(\frac{\partial g_{\mu\nu}(X)}{\partial X^\lambda}+\frac{\partial g_{\mu\lambda}(X)}{\partial X^\nu}-\frac{\partial g_{\lambda\nu}(X)}{\partial X^\mu}\right),$$ are the Christoffel symbols for $g$.

That's the EOM then: $$ \d\big(g_{\mu\lambda}(X)\ \star\, \d X^\mu\big) -\Gamma^{(g)}_{\mu\nu\lambda}\ \d X^\mu\wedge \star\, \d X^\nu = 0.$$ You can reinstate the definition of the Hodge star and of $\d$ to get a more explicit presentation: $$ \partial_\alpha\Big(\sqrt{-\det(h)}\, h^{\alpha\beta}\ g_{\mu\lambda}(X(\sigma))\;\partial_\beta X^\mu\Big) - \Gamma_{\mu\nu\lambda}^{(g)} \sqrt{-\det(h)}\, h^{\alpha\beta} \partial_{\alpha}X^\mu \partial_\beta X^\nu = 0.$$

Note that when $(M,g)=(\mathbb{R}^{D},\eta)$ this becomes the usual Polyakov EOM $$ \partial_\alpha\Big(\sqrt{-\det(h)}\, h^{\alpha\beta}\;\partial_\beta X^\mu\Big) = 0.$$

The last one is simply ($\sqrt{-\det(h)}$ times) the Laplace-Beltrami operator on $(\Sigma,h)$ acting on $X^\lambda$. Using the formula for the variation of the determinant, one can show that this is equivalent to $$ \sqrt{-\det(h)} \Big(h^{\alpha\beta} \partial_\alpha \partial_\beta - h^{\alpha\beta}{\Gamma^{(h)}}^\delta_{\alpha\beta}\partial_\delta\Big)X^\mu = 0,$$ where now ${\Gamma^{(h)}}^\cdot_{\cdot\cdot}$ are the worldsheet Christoffel symbols.

  • $\begingroup$ @Deane Which last term? $\endgroup$ Jan 20 at 17:18
  • $\begingroup$ You don't need a complicated target space for that, then... :) $\endgroup$ Jan 20 at 17:19
  • $\begingroup$ @Deane I updated the answer $\endgroup$ Jan 20 at 17:38
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    $\begingroup$ Your derivation is wrong, for a simple reason. Your index manipulations are very, very wonky. For example $\eta$ is repeated three times, $\rho$ also three times, so I stopped looking because I couldn't extract any meaning from it and decided to give you my own, correct version. I would advice that you train your index gymnastics really hard before having a go at string theory. The formula in your last comment is correct. $\endgroup$ Jan 20 at 17:48
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    $\begingroup$ @Deane That's why you should brush up your special relativity first. For example, your "$\partial_\rho\sqrt{-|h|}$" formula should read $\partial_\alpha\sqrt{-|h|} = \frac12 \sqrt{-|h|} h^{\beta\gamma} \partial_\alpha h_{\beta\gamma}$. Also I would advice you to stick to a convention for your indices. E.g. "greek indices: worldsheet; latin indices: spacetime" or "greek from the beginning: worldsheet; greek from the middle: spacetime" to minimise your confusion. $\endgroup$ Jan 20 at 19:07

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