First of all, I do recognize that I haven't studied string theory up to this point. I'm actualy just getting started with it.

So my question here is as follows: Einstein's theory of General Relativity basically says that "gravity is geometry of spacetime". That would be a very rough idea of what it is all about. The gravitational field isn't something propagating on some background, it is the background itself.

Now, string theory is said to have the potential to be the so sought theory of quantum gravity. One way to support this claim is that a massless spin 2 particle appears naturaly in the theory and this particle could be thought as the graviton. Historically it seems this particle was what motivated string theory to be used for quantum gravity and not for hadronic physics.

It is also said that string theory has the potential to be a theory of everything, unifying the four fundamental forces and all the particles in a single description.

Now, with all that said, here comes the question: it seems to me, that having in mind the basic idea of GR that gravity is the geometry of spacetime, any theory of quantum gravity should also be a quantum theory of spacetime.

Now, in string theory, as far as I know, one studies (quantum) strings, propagating in a fixed background (usually taken to be either Minkowski Spacetime or AdS spacetime) and then it ends up describing a graviton.

But how can this be a theory of quantum gravity, or even a theory of everything, if it is not a quantum theory of spacetime? In other words: spacetime is a fixed background as far as I know. Furthermore, one could interpret the graviton field as a perturbation of the background, but not all spacetimes are small perturbations of Minkowski spacetime. Actualy, I believe that at the Planck scale, where quantum gravity would be needed, it certainly wouldn't be the case that spacetime is a perturbation of Minkowski spacetime.

So my question is: how does string theory deals with this? It doesn't provide a quantum description of spacetime? If so, how can it be a true quantum gravity theory and how can it be a theory of everything?

  • 1
    $\begingroup$ I hope some strings expert answers this. The way I see it , there should be a mathematical proof that the string level at the limit of large dimensions gives the general relativity equations, similar to how newtonian gravity comes out of general relativity. for example here philsci-archive.pitt.edu/11116/1/Huggett-Vistarini.pdf $\endgroup$
    – anna v
    Aug 28, 2018 at 4:21
  • $\begingroup$ Actually, string theory is not the only competitor in the game. There is also loop quantum gravity. In the latter concept/theory space-time is quantized. There is a slight difference between loop quantum gravity and string theory: Loop quantum gravity does not claim to be the theory of everything, it is just considered of an attempt to quantize gravitation which is already complicated enough. Unfortunately my knowledge on loop quantum gravity is very little. $\endgroup$ Aug 28, 2018 at 12:51
  • $\begingroup$ @annav so your point basically is that: (1) string theory is a theory in a larger number of dimensions (I believe it is 11 in M-theory) and (2) somehow the way these strings interact lead to an effective 4-dimensional geometry which being considered on itself satisfies the equations of general relativity? $\endgroup$
    – Gold
    Aug 28, 2018 at 16:01
  • $\begingroup$ @FredericThomas the first time I've read about LQG I found its idea quite nice, but after reading more about it and discussing to people who studied it in more detail, it seems to have a bunch of internal problems: (1) it is unknown how to recover GR in the classical limit, (2) it breaks Lorentz invariance and a few others which I don't recall right now. As I said, I find this quite unfortunate, because the proposal is quite convincing. $\endgroup$
    – Gold
    Aug 28, 2018 at 16:03
  • $\begingroup$ @user1620696 yes, but as an experimentalist I wait for a theorist to supply a rigorous answer. $\endgroup$
    – anna v
    Aug 28, 2018 at 16:22

1 Answer 1


You can start by studying the Polyakov action for the string in a flat background metric $\eta_{\mu\nu}$ $$S_P = \frac{1}{4\pi \alpha '}\int d^2\sigma \sqrt{h}h^{ab}\partial_a X^\mu \partial_b X^\nu \eta_{\mu\nu}$$ and you find that the closed string contains a massless excitation identified with the graviton. A curved spacetime can be interpreted as a coherent state of gravitons (Analogy: A laser field configuration is a coherent state of photons), so with graviton excitations we should in general be considering a string in curved spacetime with metric $G_{\mu\nu}(X)$: $$S_\sigma = \frac{1}{4\pi \alpha '}\int d^2\sigma \sqrt{h}h^{ab}\partial_a X^\mu \partial_b X^\nu G_{\mu\nu}(X).$$ Now you could ask why it is justified to just change the target space metric. But if you consider a metric that is close to flat you can expand $$G_{\mu\nu}(X)=\eta_{\mu\nu}+\chi_{\mu\nu}(X)$$ where $\chi_{\mu\nu}(X)$ is a small perturbation. For the world-sheet path integral this amounts to $$Z=\int \mathcal{D}X\mathcal{D}h\exp(-S_\sigma)=\int \mathcal{D}X\mathcal{D}h\exp(-S_P)\left(1-\frac{1}{4\pi \alpha '}\int d^2\sigma \sqrt{h}h^{ab}\partial_a X^\mu \partial_b X^\nu \chi_{\mu\nu}(X)+\cdots\right)=\int \mathcal{D}X\mathcal{D}h\exp(-S_P)\left(1-V+\frac{1}{2}V^2\cdots\right)$$ where I denote $$V=\frac{1}{4\pi \alpha '}\int d^2\sigma \sqrt{h}h^{ab}\partial_a X^\mu \partial_b X^\nu \chi_{\mu\nu}(X).$$ For $\chi_{\mu\nu}=g_c \zeta_{\mu\nu}\mathrm{e}^{ik\cdot X}$ this is the vertex operator for a plane graviton wave, but more general $\chi$ could be considered. A single vertex operator $V$ would give a single graviton state but inserting the exponential as here, $\exp(V)$, corresponds to a coherent state of gravitons, which by going backwards in the above arguments corresponds to changing $$\eta_{\mu\nu} \rightarrow \eta_{\mu\nu}+\chi_{\mu\nu}=G_{\mu\nu}.$$ In conclusion: By writing down the action for a string in a curved spacetime we have implicitly inserted a coherent state of gravitons, such that the background curved metric is indeed built by quantised gravitons.

If you continue studying the sigma-model $S_\sigma$, you find that the requirement for conformal invariance (through studying the $\beta$-function) is that the target space is Ricci flat $$\mathcal{R}_{\mu\nu}=0,$$ i.e. it gives the Einstein equations in vacuum. Adding the Kalb-Ramond field and the dilation to the action will give contributions to the Einstein equations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.