To answer your question including the full supergravity, explicitly enough, one needs an amount of space not much smaller than the usual textbooks. And supergravity usually appears in their second volumes only. ;-)

However, one may divide the question to the proof of supersymmetry, and the proof of gravity in string theory. The proof that superstring theory obeys spacetime supersymmetry is a separate thing. A combination of gravity and supersymmetry immediately implies that the spacetime action has to be one of supergravity - this is pretty much by definition of supergravity, even though this verbal exercise doesn't really construct the action in components.

So the remaining question is why string theory predicts gravity.

It's because the world sheet action describes the proper area of a world sheet that propagates in a background spacetime. This spacetime may be infinitesimally changed by changing the background metric. Because the world sheet theory is scale-invariant, every infinitesimal deformation of the world sheet Lagrangian has to correspond to a marginal deformation, i.e. a dimension (1,1) tensor field.

By the state-operator correspondent, a local field of this dimension translates to an allowed, physical massless field in spacetime. Its spacetime spin is 2 because it carries the same indices as the background graviton. By adding a condensate of such closed strings in the gravitational vibrational pattern, one impacts the behavior of all other strings in the same way as if he deformed the background. This follows from the state-operator correspondence.

So the principle of equivalence is automatically satisfied: the graviton closed strings behave just like perturbations of the background, and they act on the motion of all objects and fields in the same way. The spacetime geometry itself follows Einstein's equations because they're the only diff-invariant locally Lorentz-invariant equations that they could satisfy - whose linear part is a wave equation.

More explicitly, one may show that one gets the Ricci tensor in spacetime because it's a consistency condition for the allowed backgrounds where strings may propagate. The string dynamics may only be defined by scale-invariant world sheet theories, and the change of the world sheet action with respect to the scale - also known as the beta-function (for each coupling constant) - has to vanish as a consequence.

For the worldsheet, the metric tensor at each spacetime point is a coupling constant and the beta-function may be calculated from one-loop graphs in the world sheet theory, and it ends up being the Ricci tensor, roughly speaking. It follows that the background has to satisfy the right equations of general relativity at long distances.

The state-operator correspondence guarantees that the closed-string excitations also interact according to the same equations because the closed strings are equivalent to infinitesimal deformations of the background.

Supersymmetry is added as a lot of extra fermionic stuff and leads to supergravity in spacetime - which is still just gravity coupled to other matter fields in a way that preserves spacetime supersymmetry. See also

http://motls.blogspot.com/2007/05/why-are-there-gravitons-in-string.html

This post imported from StackExchange Physics at 2014-04-01 16:53 (UCT), posted by SE-user Luboš Motl