# How do Einstein's field equations come out of string theory?

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?

• 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. Nov 21, 2012 at 19:11
• @John Rennie for example, how to derive this relation: $ds^2=g_{\mu\nu}x^{\mu}x^{\nu}$ from string theory?
– Neo
Nov 21, 2012 at 19:20
• 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. Nov 21, 2012 at 19:32
• 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. Nov 21, 2012 at 19:35
• @Luboš Motl $g_{\mu\nu}(X^{\alpha})$ this is very similar. what means $(X^{\alpha})$?
– Neo
Nov 21, 2012 at 20:52

$$\mbox ds^2=g_{\mu\nu}\mbox dx^{\mu}\mbox dx^{\nu}$$ is a definitional fact from Riemannian geometry, and has nothing to do with gravity. The "physics" of General Relativity is contained in the Einstein Field equations $$G_{\mu\nu}=8\pi T_{\mu\nu}$$, or equivalently the Einstein-Hilbert action $$\mathcal{L}=\frac1{16\pi}R$$.

Deriving these results from the Polyakov action is hard, but there is a simpler standard approach that you'll find in many textbooks. In string theory, the Dilaton couples to the worldsheet

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

The breaking of conformal symmetry in this action can be summarized by 3 functions known as the beta functions. In Type IIB string theory, the beta functions are:

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

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

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

Setting these functions to zero (i.e. to require conformal symmetry, expecting to obtain the vacuum Einstein-Field Equations):

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

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

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

The first of these equations is a corrected form of the vacuum EFE, and the remaining equations represent analogous equations for other fields.

• 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). Aug 22, 2013 at 22:58