The classical theory of spacetime geometry that we call gravity is described at its core by the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy in spacetime.

For example: $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ is an important concept in General Relativity.

Mathematically, how do the Einstein's equations come out of string theory?

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    $\begingroup$ It's not exactly a duplicate, but Luboš' answer to physics.stackexchange.com/q/44732 is as close as you'll get without the answer turning into a book on string theory. $\endgroup$ – John Rennie Nov 21 '12 at 19:11
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    $\begingroup$ @John Rennie for example, how to derive this relation: $ds^2=g_{\mu\nu}x^{\mu}x^{\nu}$ from string theory? $\endgroup$ – Neo Nov 21 '12 at 19:20
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    $\begingroup$ Dear Neo, the question "how to derive $ds^2=g\cdot x\cdot x$" is meaningless because one may always say that it's a definition of $ds^2$, whether one talks about string theory or not. One could ask why this expression is constant under Lorentz transformation, but it's also true by the definition of the Lorentz group, or because of basic maths, or one could ask why string theory is invariant under this group, which is easily checked because its defining objects such as action are nicely contracting the spacetime vector indices. $\endgroup$ – Luboš Motl Nov 21 '12 at 19:32
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    $\begingroup$ As John says, if you click at the previous question, you will learn that Einstein's equations arise either from effective action one may derive from scattering amplitudes, or from the vanishing of the beta-functions for the metric tensor functions which are "infinitely many coupling constants" of the world sheet theory and the world sheet theory must be conformal (scale-invariant). Explaining all these things with everything one needs to technically understand it is pretty much equivalent to teaching you introduction to string theory which is a 1-semester course, not 1 question on Stack Exc. $\endgroup$ – Luboš Motl Nov 21 '12 at 19:35
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    $\begingroup$ @Luboš Motl $g_{\mu\nu}(X^{\alpha})$ this is very similar. what means $(X^{\alpha})$? $\endgroup$ – Neo Nov 21 '12 at 20:52

Firstly. $\mbox ds^2=g_{\mu\nu}\mbox dx^{\mu}\mbox dx^{\nu}$ is not specific to General Relativity. It can be thought either as the definition of $g_{\mu\nu}$ or of $\mbox ds^2$. It is from Riemannian Geometry, in general.

What you could ask is "How does one derive the Einstein Field Equation $G_{\mu\nu}=\frac{8\pi G}{c_0^4}T_{\mu\nu}$ or the Einstein-Hilbert Lagrangian Density $\mathcal L = \frac{c_0^4}{16\pi G}R$ from String Theory?". So, I'll consider that your question and answer it that way.

Firstly, see the answers (which includes mine) at :

The General Relativity from String Theory Point of View

So, deriving these things from the Polyakov action (since gravitons are bosons) is quite hard. Oh, no!!!!!! Well, fortunately, there is a veeeery easy method. Called the Beta function. In string theory, the Dilaton couples to the worldsheet

$$S_\Phi = \frac1{4\pi} \int d^2 \sigma \sqrt{\pm h} R \Phi(X)$$

The $\pm$ is supposed to indicate that it depends on convention. Notice the terms inside the action integral? Call the conformal police! The dilaton field broke Conformal symmetry!!!

We can't let that happen! Luckily, this breakage of this law has been summarised in 3 functions, called Beta functions. There are 3 beta functions in say, Type IIB string theory:

$${\beta _{\mu \nu }}\left( g \right) = \ell _P^2\left( {{R_{\mu \nu }} + 2{\nabla _\mu }{\nabla _\nu }\Phi - {H_{\mu \nu \lambda \kappa }}H_\nu ^{\lambda \kappa }} \right)$$

$$ {\beta _{\mu \nu }}\left( F \right) = \frac{{\ell _P^2}}{2}{\nabla ^\lambda }{H_{\lambda \mu \nu }} $$

$$ \beta \left( \Phi \right) = \ell _P^2\left( { - \frac{1}{2}{\nabla _\mu }{\nabla _\nu }\Phi + {\nabla _\mu }\Phi {\nabla ^\mu }\Phi - \frac{1}{{24}}{H_{\mu \nu \lambda }}{H^{\mu \nu \lambda }}} \right) $$

Where $\ell_P$ is the string length (you may want to confuse this with the string length. If so, please do so.) . If we just kill these breakages, i.e. set the breakages equal to 0: . : . : .

$${{R_{\mu \nu }} + 2{\nabla _\mu }{\nabla _\nu }\Phi - {H_{\mu \nu \lambda \kappa }}H_\nu ^{\lambda \kappa }} = 0. $$

$${\nabla ^\lambda }{H_{\lambda \mu \nu }} = 0 . $$

$$ { - \frac{1}{2}{\nabla _\mu }{\nabla _\nu }\Phi + {\nabla _\mu }\Phi {\nabla ^\mu }\Phi - \frac{1}{{24}}{H_{\mu \nu \lambda }}{H^{\mu \nu \lambda }}} = 0 . $$

Look at the first one, because that was killed (I'm a murderered!) to ensure conformal invariance for gravity (which explains $\beta_{ \mu\nu }\left(G \right)$). Now, isn't this just the EFE with the stringy corrections from the Dilaton field?! Q.E.D.

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  • $\begingroup$ If you wish to derive the beta-function, use Riemann normal coordinates, this is the best way. It's described on Wikipedia, then the beta-function calculation is a piece of cake (relatively, once you figure out what you're doing exactly). $\endgroup$ – Ron Maimon Aug 22 '13 at 22:58

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