This article

K. Pardo, et. al., Limits on the number of spacetime dimensions from GW170817, Journal of Cosmology and Astroparticle Physics, Vol. 2018, 2018.

which was published recently in JCAP states that they put constraints on the number of extra non-compact dimensions by looking at the difference between the Electromagnetic luminosity distance $d_L^{EM}$ and the gravitational luminosity distance $d_L^{GW}$. The approach makes sense given the assumption that gravitational waves "leak" into the extra dimensions and electromagnetic waves do not. If this assumption is true, then a given source would have $d_L^{GW}>d_L^{EM}$ since the gravitational wave signal would be weaker than expected. By looking at the ratio of $d_L^{GW}/d_L^{EM}$ one could find constraints on "how much" of the gravitational waves is "leaking". What is the basis of this assumption though? Why can only gravitational waves leak into extra non-compact dimensions and electromagnetic waves can't?


1 Answer 1


The analysis is model dependent, and in a string theory setting would correspond to a brane-world scenario where the non-gravitational fields (open strings) live on a 3+1 dimensional brane. That is, the non-gravitational fields are confined to the 3+1 dimensional space of the brane, while gravity (closed strings) are free to propagate in the bulk (the extra dimensions).

Examples of models with extra dimensions where the non-gravitational sector is assumed to be confined to the 3+1 dimensions are for example the Randall–Sundrum model, the ADD model and the Dvali–Gabadadze–Porrati braneworld.

  • $\begingroup$ Pardon my ignorance, I know very little about string theory, but what is it about gravity in these models, then, that makes it "closed strings" and free to propagate in the bulk but EM fields are "open strings" and confined to live on the 3+1 D brane? For example, the differentiator between EM fields and Gravitational fields in GR would be in its application of the equivalence principle which equates gravity (and not EM fields) to space-time curvature. When it comes down to it, is it also the equivalence principle that differentiates gravity from EM in string theories? $\endgroup$
    – enumaris
    Sep 16, 2018 at 18:28
  • $\begingroup$ @enumaris They are different excitations of the string. For closed strings you find that the first excited state contains a graviton. When studying the open string you have to define boundary conditions for the ends of the string: Neumann (N) or Dirichlet (D). Choosing NN boundary conditions along coordinates $X_\mu$ for $\mu =0, \ldots, p$ and DD for the remaining defines for you a (p+1)-dim. hypersurface of spacetime on which open strings can end. The excitations of the open string along this brane can be found to contain gauge fields (like EM), i.e. the gauge theory lives on the brane. $\endgroup$
    – Sparticle
    Sep 16, 2018 at 20:40
  • $\begingroup$ Thanks for this answer. I think it answers the question of why closed strings can "leak" while open strings are confined to the brane. I was wondering though, the other question, why are gravitons closed strings while other gauge bosons are open strings? $\endgroup$
    – enumaris
    Sep 17, 2018 at 16:25
  • $\begingroup$ @enumaris You would have to study the quantisation of strings, closed and open, to really appreciate how the different states arise in different sectors. It is too involved to answer here as a comment. In a nutshell, doing the calculations you simply get the massless symmetric traceless representation (graviton) as an excitation of the closed string, and the vector boson as an excitation of the open string. $\endgroup$
    – Sparticle
    Sep 17, 2018 at 20:53
  • $\begingroup$ Stated differently, the graviton is supposed to be a spin 2 particle, whereas gauge bosons should be spin 1. When one quantizes string theory, one finds a massless spin 2 excitation in the closed string sector, and a massless spin 1 excitation in the open string sector. This is essentially because the closed string has two sets of vibrational modes (right and left movers), but the open string has only one (there, the right moving modes are reflected into the left moving ones, so their excitations are the same). $\endgroup$
    – Stijn B.
    Sep 28, 2018 at 8:12

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.