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Pedagogical explanations of critical dimensions of string theories

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Once you understand the formalism, I think it's clearest to say the critical dimension of the space-time arises because we need to cancel the central charge of the (super)conformal ghosts on the worldsheet.

But suppose I need to explain this to a person who only knows very basic QFT. How would you explain the concept of the critical dimension?

This post imported from StackExchange Physics at 2014-03-26 12:54 (UCT), posted by SE-user Yuji
asked Feb 11, 2011 in Theoretical Physics by Yuji (1,390 points) [ no revision ]

1 Answer

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There are about 10 "significantly different" ways of calculating the critical dimension but all of them require some maths. n For the case of bosonic string theory, one may show that the $bc$ system carries the central charge $-3k^2+1$ for $k=3$ which is $-26$, so it has to be canceled by 26 bosons.

Alternatively, one may define the light-cone gauge string. Only $D-2$ transverse dimensions produce oscillators. There are oscillators $$\alpha_n^i$$ where $i$ has $D-2$ different values and $n$ is an integer. The squared mass (in spacetime) of the string i.e. particle is given by the total excitation of all the harmonic oscillators $$\sum_{n>0,i} \alpha_{-n}^i \alpha^i_n$$ The oscillators $\alpha_n$ are normalized as $\sqrt{n}\hat a$ for a normal annihilation operator of the harmonic oscillator, so the excitations are weighed by $n$. Consequently, the zero-point energy is $$\frac{D-2}{2}\sum_{n=1}^\infty n$$ The sum of positive integers is equal to $-1/12$ - by the zeta-function regularization or any equally good one (the correct final result of the sum has been known for a century or more). The first excited state - excited by $\alpha_{-1}^i$ - adds one to the energy and it is a spacetime vector. However, it only has $D-2$ components rather than $D-1$ components, so it must correspond to a massless particle. Consequently, $$\frac{D-2}{2} \left(-\frac{1}{12}\right)+1 = 0$$ has to hold. It follows that $D-2=24$, $D=26$.

The calculations of the superstring critical dimension, $D=10$, have to account for the world sheet fermions as well (which can be added via several equivalent machineries, either as spacetime vectors, RNS, or spacetime spinors, GS). Alternatively, one may adopt a spacetime perspective and explain why 10 and 11 are the right dimensions in which maximally supersymmetric string theory and M-theory may live.

In some sense, a basic knowledge of QFT is enough to understand those calculations - but less than the basic knowledge of QFT is just not enough. As far as I know, there is no honest "high school level" derivation of the critical dimension - even though the final result may be understood by the high school kids.

This post imported from StackExchange Physics at 2014-03-26 12:54 (UCT), posted by SE-user Luboš Motl
answered Feb 11, 2011 by Luboš Motl (10,178 points) [ no revision ]
Thanks, could you list ten different ways, so that I can learn?

This post imported from StackExchange Physics at 2014-03-26 12:54 (UCT), posted by SE-user Yuji
Dear Yuji, sorry, this is too time-consuming because I would probably have to explain the objects. Check Joe Polchinski's textbook that derives the critical dimension in 7 ways - but you won't find a list of the ways, anyway. ;-) This is just not how physics works. There is no precise number "10" for the number of ways, and there's no precise way to say whether two methods are really "the same" or not. I listed two ways to get $D=26$ in the comment above; the historically first appearance of $D=26$ was actually in a more complex context - in the unitarity of loop amplitudes.

This post imported from StackExchange Physics at 2014-03-26 12:54 (UCT), posted by SE-user Luboš Motl
OK, I guess I need to go back to Polchinski's bible... thanks anyways

This post imported from StackExchange Physics at 2014-03-26 12:54 (UCT), posted by SE-user Yuji

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