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  Critical radius in heterotic string theory

+ 1 like - 0 dislike

If I have understood the basics of heterotic strings correctly, one requires a critical radius of compactified dimension to reconcile the apparent difference in spacetime dimension of left and right movers,

$$ R=\sqrt{\alpha} $$

Why would the radius of the compactified 16 dimensional torus be related to the string tension/length etc in such a way? They seem completely unrelated to me.

This post imported from StackExchange Physics at 2014-03-17 06:13 (UCT), posted by SE-user innisfree

asked Nov 22, 2013 in Theoretical Physics by innisfree (295 points) [ revision history ]
edited Apr 19, 2014 by dimension10

1 Answer

+ 2 like - 0 dislike

The 16 toroidal dimensions have a stringy radius because it has to be self-dual under T-duality, $$ R \to \frac{\alpha'}{R}$$ This is needed for these 16 dimensions to be purely left-moving. A fast semi-heuristic way to see it is that the left-moving dimensions obey $$\partial X = 0, \quad \partial_\sigma X = \partial_\tau X$$ If you integrate the latter form of the equation over $\sigma$, you will get $$ \alpha' p = w$$ The 16-momentum is equal to the 16-winding for all the states, in string units. But the momentum and winding belong to lattices that are dual ("inversely proportional") to one another because the momentum spacing goes like $1/R$ while the winding is a multiple of $2\pi R$. So the allowed radius has to be $\sqrt{\alpha'}$ in certain conventions where the factors of $2,\pi$ are properly accounted for.

This post imported from StackExchange Physics at 2014-03-17 06:13 (UCT), posted by SE-user Luboš Motl
answered Dec 6, 2013 by Luboš Motl (10,278 points) [ no revision ]
Thanks for taking the time to look at this!

This post imported from StackExchange Physics at 2014-03-17 06:13 (UCT), posted by SE-user innisfree

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