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I know that current string theories state that there are 10, 11, or 26 spacetime dimensions in superstring theory, M-theory, and bosonic string theory, respectively. But when I looked up why those numbers were chosen and not, say, 35, I could not find an answer that suited me.

As I am not an expert in string theory, can you please explain to me why (if possible in easier terms) those specific answers were chosen in regard to the number of dimensions?

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marked as duplicate by ACuriousMind Nov 16 '16 at 16:14

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  • $\begingroup$ I don't really know anything about string theory, even though I have wondered the same thing. So since I'm no authority I'm making a comment rather than an answer. A string is an infinitesimally small thing that is characterized by its vibrations. To account for all of the physical characteristics of the universe, strings would have to vibrate in multiple dimensions. So there are theories that try to make a TOE with vibrations in 10 dimensions, another in 11 dimensions and another in 26 dimensions. Maybe in actuality there are an infinite number of dimensions, most of which are not used. $\endgroup$ – Howard Miller Nov 15 '16 at 22:19
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  • $\begingroup$ There are different explanations for the numbers 10, 11, 26 all of which come from some theoretical requirement of internal self-consistency of the mathematical theory. You will need some background in physics and mathematics to truly understand. What is your current background? $\endgroup$ – Prahar Nov 15 '16 at 23:29
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In bosonic string theory, after canonical quantisation, we can construct particle states from the creation operators $\tilde{\alpha}^i_{-1}$ and $\alpha^j_{-1}$, as $\tilde{\alpha}^i_{-1} \alpha^j_{-1} |0; p\rangle$, which turn out to each have a mass of,

$$M^2 = \frac{4}{\alpha'}\left( 1- \frac{d-2}{24}\right)$$

where $d$ is the space-time dimension and $\alpha'$ is the Regge slope. Why is this a problem? In short, Wigner's classification of representations of the Poincaré group says any massive particle must form a representation of $SO(d-1)$ in $\mathbb{R}^{1,d-1}$. Without going into the entire canonical quantisation, the particles described have $(d-2)^2$ states. However, if these particles happen to be massless, they are permitted and have fewer internal states than massive particles, just like how a photon in four dimensions has two degrees of freedom - its two polarisations.

Thus, for them to be massless and preserve Lorentz symmetry, we must require $M^2=0$, so $d=26.$


There is another way to view the reason for $d=26$ in bosonic string theory. When examining the CFT given by the Polyakov action, one of the conditions of a CFT one can derive is,

$$T^\alpha_\alpha = 0$$

that is, the trace of the stress-energy tensor for the theory vanishes. However, when quantising the theory, it turns out that,

$$\langle T^\alpha_\alpha\rangle = -\frac{c}{12}R$$

where $c$ is the central charge and $R$ is the Ricci scalar of the curved background of our theory. This is known as an anomaly, and it arises in 'simpler' theories, like the chiral anomaly of quantum electrodynamics.

Now, when quantising the CFT using the path integral, ghost particles necessarily arise, just like they do when quantising non-Abelian gauge field theories. It turns this ghost system has $c = -26$. So, for the whole theory to have $c = 0$ and thus keep Weyl symmetry, we need to add $26$ scalar fields which each contribute $c = 1$.


Although you've asked for a non-technical description, the reality is these conditions stem from specific mathematical details of the theory, and more specifically from the symmetries they must possess.

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  • $\begingroup$ Nice. But. In order for this to truly be an explanation rather than merely shifting the authoritative statement back one level, the reader really has to be able to follow the reasoning behind everything taken as a given here. Canonical quantization. Representations of the group. Why massless particles get to ignore the internal state rule. And so on. So that the addendum under the final rule is the main content of the answer for the lay reader. $\endgroup$ – dmckee Nov 16 '16 at 2:39
  • $\begingroup$ @dmckee Yeah, I honestly didn't know how to present this because the right way to get to $d=26$ or $d=10$ for the superstring, is to go through the entire derivation, which spans many pages, and I obviously couldn't present that here. It's hard to strike a balance between too little and too much information for this question. $\endgroup$ – JamalS Nov 16 '16 at 8:17

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