I'm a physicist, but my knowledge of string theory is extremely minimal. My naive conceptual understanding is that the vacuum is modeled as a certain topology (and geometry?) for the spacetime, and fundamental particles are explained as excitations of strings that live on this background. E.g., a certain vibrational state would be the photon, some other state would be the graviton, and so on. The actual spectrum of such vibrations is presumably something we can't calculate, because we don't know the topology of the background (i.e., which possibility it is in the string landscape).

If this is at least qualitatively correct, then why aren't there infinitely many such vibrational states, which would appear at ordinary energy scales as infinitely many elementary particles?

  • 2
    $\begingroup$ related: Does String Theory predict a particle with twice the mass of the electron?. $\endgroup$ Commented Nov 7, 2019 at 16:44
  • 3
    $\begingroup$ String theory does predict an infinite number of particles, but all except the lowest modes have masses of order of the Planck mass. $\endgroup$ Commented Nov 7, 2019 at 17:49
  • $\begingroup$ @JohnRennie: I see, that sort of makes sense. Maybe that should be an answer. But that leaves me wondering why the lowest mode doesn't also have a mass on the order of the Planck mass. I would think that it would be like a simple harmonic oscillator with $E_n=(n+1/2)\omega_0$, so that $E_0$ would be on the order of $\omega_0$. $\endgroup$
    – user4552
    Commented Nov 7, 2019 at 23:15
  • 1
    $\begingroup$ The lowest massive modes are at the Planck mass. There are also massless modes that get masses through symmetry breaking. (To be annoying and cite a whole book: see Polchinkski.) $\endgroup$
    – d_b
    Commented Nov 8, 2019 at 3:24
  • $\begingroup$ No, I was agreeing with him. The small masses of quarks and leptons come from Yukawa coupling between fermions and Higgs + spontaneous symmetry breaking. $\endgroup$
    – d_b
    Commented Nov 8, 2019 at 4:35

1 Answer 1


For simplicity take the bosonic string theory in 26 dimensions. When you quantise the open and closed string you find excitations (states) of the string at any level $N$ with masses \begin{equation} M^2_\mathrm{open}=\frac{1}{\alpha '}\left(N-1\right),\qquad M^2_\mathrm{closed}=\frac{4}{\alpha '}\left(N-1\right). \end{equation} As $N$ takes on any non-negative value, you see that there is indeed an infinite tower of states. The tachyonic state (negative mass) at $N=0$ is a sickness of the purely bosonic string, which goes away for the superstring. At level $N=1$ you find the massless excitations (in 26 dimensions), and from the next level all states are massive. The mass is determined by the pre-factor. You can regard $\alpha '$ as the only free parameter of string theory, and it relates to the tension, $T$, of the string, which is expected to be set by the string scale, which is slightly below the Planck scale \begin{equation} \frac{1}{\alpha '}= 2\pi T \lesssim M_\mathrm{Pl}^2 = \left(10^{19} \, \mathrm{GeV} \right)^2 . \end{equation} So in conclusion, the infinite tower of massive excited states have masses at the order of the Planck scale, which means they are unobservable, and there is only a finite number of massless excitations. The same thing goes for the superstring.

  • $\begingroup$ This seems helpful to me. Maybe the downvoter could explain whether there was something they disagreed with. It would be nice to have some explanation for the non-specialist like me of we why get $M^2\propto (N-1)$, but maybe I should research that myself of make it a separate question. d_b says in a comment that "The small masses of quarks and leptons come from Yukawa coupling between fermions and Higgs + spontaneous symmetry breaking." If this is correct, then adding something like this might improve the answer. $\endgroup$
    – user4552
    Commented Nov 8, 2019 at 16:36
  • $\begingroup$ @BenCrowell The mass spectrum, $M^2\propto (N-1)$, is a standard textbook derivation when quantising the string, and I wouldn't know of a convincing way to show it quickly unfortunately. Regarding small masses: At the end of the day, the effective low energy description of string theory, should be described by the Standard Model (+more). So after compactifications etc., massless modes do acquire a mass through the Higgs mechanism. This has a "stringy" description in terms of strings that acquire a mass by stretching between branes that are separated. $\endgroup$
    – Sparticle
    Commented Nov 9, 2019 at 16:15
  • $\begingroup$ @BenCrowell It doesn't change the conclusion to your question though, that the quantisation of the string gives a finite number of massless, or light through Higgsing, modes and an infinite number of particles that are just too heavy to be excited. $\endgroup$
    – Sparticle
    Commented Nov 9, 2019 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.