String Theory is formulated in 10 or 11 (or 26?) dimensions where it is assumed that all of the space dimensions except for 3 (large) space dimensions and 1 time dimension are a compact manifold with very small dimensions. The question is what is assumed about the curvature of the large 3 space and 1 time dimensions? If these dimensions are assumed to be flat, then how is String Theory ever able to reproduce the equations of General Relativity which require curved space time in the presence of mass-energy (of course, the actual source term for General Relativity is the stress energy tensor).

On the other hand if String Theory is formulated on a general curved space-time with an unknown metric (usually signified by $g_{\mu\nu}$) how do the equations of General Relativity that puts constraints on $g_{\mu\nu}$ arise from string theory?

It is well known that General Relativity requires a spin-2 massless particle as the "force mediation" particle (similar to the photon as the spin-1 massless force mediation particle of electromagnetism). It is also well know that String Theory can accommodate the purported spin-2 massless particle as the oscillation of a closed string. But how does this graviton particle relate to the curvature of the large dimensions of space-time?

I am aware that " How does String Theory predict Gravity? " is somewhat similar to this question, but I do not think it actually contains an answers to this question so please don't mark it as a duplicate question. I would especially appreciate an answer that could be understood by a non-theoretical ("String Theory") physicist - hopefully the answer would be at a level higher than a popular non-mathematical explanation. In other words, assume the reader of the answer understands General Relativity and particle physics, but not String Theory.

Update from Comment to Clarify: If you start with flat space then $g_{\mu\nu}$ isn't the metric tensor since you assumed flat space. If you start with arbitrarily curved space why would and how could you prove that the components of the graviton give you the metric tensor? I am interested in the strongly curved space case since that is where GR differs the most from Newtonian gravity. In flat space you could sort of consider weak Newtonian gravity to be the result of the exchange of massless spin-2 particles. But strong gravity needs actual space curvature to be equivalent to GR.

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    $\begingroup$ Considering quantum gravity to be a gauge theory too, the graviton (which can not be avoided in ST) is described by the metric tensor field which gives the curvature of space time and couples to the energy-momentum tensor which takes the role of the charge. That is how I imagine it to work, at least in situations with not too strong gravitational fields. $\endgroup$
    – Dilaton
    Nov 21, 2012 at 2:57
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    $\begingroup$ @dilaton, I understand the graviton cannot be avoided. Your comment maybe kinda makes sense. But if you start with flat space then $g_{\mu\nu}$ isn't the metric since you assumed flat space. If you started in arbitrary curved space why would and how could you prove that the components of the graviton are actually the metric tensor? I guess I am interested in the strongly curved space since that is where GR differs the most from Newtonian gravity. In flat space you could consider weak Newtonian gravity from exchange of massless particles. But strong gravity needs space curvature. $\endgroup$
    – FrankH
    Nov 21, 2012 at 4:01
  • $\begingroup$ This doesn't really have anything to do with string theory. Your question is more a long the lines of how can a linearized field theory look like a full nonlinear geometric theory? The answer and procedure takes quite a few pages to write down, but mathematically its actually very much like what happens with nonabelian gauge theory where you can ask a similar question. eg "how can gluons be quanta". In the case of gravity, unfortunately the whole story has been obfuscated by a group of people who confuse themselves needlessly (possibly for historical reasons) but it really is quite simple $\endgroup$
    – Columbia
    Nov 21, 2012 at 6:58
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    $\begingroup$ @FrankH +1 take a look to my question: What is Classical limit of String Theory? $\endgroup$
    – Neo
    Nov 21, 2012 at 7:43
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2 Answers 2


String theory may be considered as a framework to calculate scattering amplitudes (or other physically meaningful, gauge-invariant quantities) around a flat background; or any curved background (possibly equipped with nonzero values of other fields) that solves the equations of motion. The curvature of spacetime is physically equivalent to a coherent state (condensate) of closed strings whose internal degrees of freedom are found in the graviton eigenstates and whose zero modes and polarizations describe the detailed profile $g_{\mu\nu}(X^\alpha)$.

Einstein's equations arise as equations for the vanishing of the beta-functions – derivatives of the (continuously infinitely many) world sheet coupling constants $g_{\mu\nu}(X^\alpha)$ with respect to the world sheet renormalization scale – which is needed for the scaling conformal symmetry of the world sheet (including the quantum corrections), a part of the gauge symmetry constraints of the world sheet theory. Equivalently, one may realize that the closed strings are quanta of a field and calculate their interactions in an effective action from their scattering amplitudes at any fixed background. The answer is, once again, that the low-energy action is the action of general relativity; and the diffeomorphism symmetry is actually exact. It is not a surprise that the two methods produce the same answer; it is guaranteed by the state-operator correspondence, a mathematical fact about conformal field theories (such as the theory on the string world sheet).

The relationship between the spacetime curvature and the graviton mode of the closed string is that the former is the condensate of the latter. They're the same thing. They're provably the same thing. Adding closed string excitations to a background is the only way to change the geometry (and curvature) of this background. (This is true for all of other physical properties; everything is made out of strings in string theory.) On the contrary, when we add closed strings in the graviton mode to a state of the spacetime, their effect on other gravitons and all other particles is physically indistinguishable from a modification of the background geometry. Adjustment of the number and state of closed strings in the graviton mode is the right and only way to change the background geometry. See also


Let me be a more mathematical here. The world sheet theory in a general background is given by the action $$ S = \int d^2\sigma\,g_{\mu\nu}(X^\alpha(\sigma)) \partial_\alpha X^\mu(\sigma)\partial^\alpha X^\nu(\sigma) $$ It is a modified Klein-Gordon action for 10 (superstring) or 26 (bosonic string theory) scalar fields in 1+1 dimensions. The functions $g_{\mu\nu}(X^\alpha)$ define the detailed theory; they play the role of the coupling constants. The world sheet metric may always be (locally) put to the flat form, by a combination of the 2D diffeomorphisms and Weyl scalings.

Now, the scattering amplitudes in (perturbative) string theory are calculated as $$ A = \int {\mathcal D} h_{\alpha\beta}\cdots \exp(-S)\prod_{i=1}^n \int d^2\sigma V_i $$ We integrate over all metrics on the world sheet, add the usual $\exp(-S)$ dependence on the world sheet action (Euclideanized, to make it mathematically convenient by a continuation), and insert $n$ "vertex operators" $V_i$, integrated over the world sheet, corresponding to the external states.

The key thing for your question is that the vertex operator for a graviton has the form $$V_{\rm graviton} = \epsilon_{\mu\nu}\partial_\alpha X^\mu (\sigma)\partial^\alpha X^\nu(\sigma)\cdot \exp(ik\cdot X(\sigma)).$$ The exponential, the plane wave, represents (the basis for) the most general dependence of the wave function on the spacetime, $\epsilon$ is the polarization tensor, and each of the two $\partial_\alpha X^\mu(\sigma)$ factors arises from one excitation $\alpha_{-1}^\mu$ of the closed string (or with a tilde) above the tachyonic ground state. (It's similar for the superstring but the tachyon is removed from the physical spectrum.)

Because of these two derivatives of $X^\mu$, the vertex operator has the same form as the world sheet Lagrangian (kinetic term) itself, with a more general background metric. So if we insert this graviton into a scattering process (in a coherent state, so that it is exponentiated), it has exactly the same effect as if we modify the integrand by changing the factor $\exp(-S)$ by modifying the "background metric" coupling constants that $S$ depends upon.

So the addition of the closed string external states to the scattering process is equivalent to not adding them but starting with a modified classical background. Whether we include the factor into $\exp(-S)$ or into $\prod V_i$ is a matter of bookkeeping – it is the question which part of the fields is considered background and which part is a perturbation of the background. However, the dynamics of string theory is background-independent in this sense. The total space of possible states, and their evolution, is independent of our choice of the background. By adding perturbations, in this case physical gravitons, we may always change any allowed background to any other allowed background.

We always need some vertex operators $V_i$, in order to build the "Fock space" of possible states with particles – not all states are "coherent", after all. However, you could try to realize the opposite extreme attitude, namely to move "all the factors", including those from $\exp(-S)$, from the action part to the vertex operators. Such a formulation of string theory would have no classical background, just the string interactions. It's somewhat singular but it's possible to formulate string theory in this way, at least in the cubic string field theory (for open strings). It's called the "background-independent formulation of the string field theory": instead of the general $\int\Psi*Q\Psi+\Psi*\Psi*\Psi$ quadratic-and-cubic action, we may take the action of string field theory to be just $\int\Psi*\Psi*\Psi$ and the quadratic term (with all the kinetic terms that know about the background spacetime geometry) may be generated if the string field $\Psi$ has a vacuum condensate. Well, it's a sort of a singular one, an excitation of the "identity string field", but at least formally, it's possible: the whole spacetime may be generated purely out of stringy interactions (the cubic term), with no background geometry to start with.

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    $\begingroup$ Wow, this is such a very nice answer and I will have to read the TRF article linked to too :-). In addidion to nicely explain how gravity really works, this post efficiently proves any claims, often thrown around in popular science magazines that ST is no good because it is not background independent etc, wrong. I like this answer a lot, +1 and stared $\endgroup$
    – Dilaton
    Nov 21, 2012 at 8:10
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    $\begingroup$ Thank for your interest, folks! Lurscher: in the background-independent formulation, the "physical" vacuum state with the ordinary regular metric tensor etc. is a condensate of many strings in a very complicated state. The only state without any particle excitations is the state with the values of all fields set to zero, i.e. $\Psi=0$ in this case. Everything else has some quanta, by definition. They're zero-momentum quanta but they're there. So $\Psi=0$ in the background-independent variables is a different state - different vacuum, if you wish - than the normal one. $\endgroup$ Nov 21, 2012 at 15:51
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    $\begingroup$ The string field theory action is a functional integral over all shapes of half-strings, $\int {\mathcal D}X(\sigma)$. It's surely not just an integral over the spacetime, it also includes all the integrals over the nonzero modes. Please, this is not a blog entry about string field theory and a full introduction to basics of string theory doesn't belong to any answers on SE, anyway. Please find some intro to string field theory if you want to understand it. $\endgroup$ Nov 21, 2012 at 15:53
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    $\begingroup$ Interesting, It will take me a long time digest every line of it. As I understand from what I read, varying the background metric in the path integral has no relevance in string theory. Is it so? $\endgroup$
    – Prathyush
    Nov 21, 2012 at 17:35
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    $\begingroup$ Right, Prathyush. In the world sheet theory, the metric tensor as a function of the spacetime is a set of coupling constants, not dynamical fields or degrees of freedom, so we don't vary those things like in GR. At the same time, when we change them to another nearby solution, we get an equivalent formulation of string theory. $\endgroup$ Nov 22, 2012 at 5:38

First a sub-question: If Einstein space-time "arise[s] as equations for the vanishing of the beta-functions – derivatives of the (continuously infinitely many) world sheet coupling constants gμν(Xα) with respect to the world sheet renormalization scale – which is needed for the scaling conformal symmetry of the world sheet (including the quantum corrections), a part of the gauge symmetry constraints of the world sheet theory," then how does one arrive at various closed shapes for the universe?

Second, my answer: It is fairly easy to produce an exact local gravitational field as an emergent curved space-time on a flat background. Takes only a page or two. See https://link.springer.com/article/10.1140/epjp/i2018-11983-2 However, this, like in my sub-question above, leaves something to be desired in forming closed cosmological space-times.


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