T duality under a small fluctuation of the compact dimension

Question

How do small perturbations around the compact dimension affect T duality. What happens if I chose a compactification of the nature $r+\delta r$. And what keeps the compact dimension stable, i.e from expanding into a large dimension or imploding into itself.

This post imported from StackExchange Physics at 2015-03-23 11:10 (UTC), posted by SE-user Prathyush

Answer 1

T-duality says that the radius $r$ is equivalent to $\alpha' / r$. So $r+\delta r$ is also equivalent to $\alpha'/(r+\delta r)$, too. If the radius fluctuates, so does its T-dual radius.

The radius itself, usefully written as $\sqrt{\alpha'}\exp(\phi_R)$, is a "modulus", a scalar field that has no potential (i.e. all conceivable values are equally allowed: this implies that the scalar field is massless around any point and produces new long-range forces that would invalidate the equivalence principle - the experimentally verified principle that all objects accelerate by the same acceleration in a gravitational field) in non-realistic vacua but a potential is generated in realistic ("stabilized") vacua. In the latter vacua, $r=r_0$ chooses a particular value for which $V'(r_0)=0$, the potential has to be minimized.

One may still describe the same point using $\tilde r_0 =\alpha' / r_0$ but we usually pick the larger among these two T-dual values because large compactification radii (longer than the string length) are those in which string theory agrees with the low-energy effective field theory more directly (the excited strings may be neglected for many purposes).

What are the physical reasons that generate such a potential is an extensive technical topic that has to be solved mostly independently in separate classes of the string vacua. Quite generally, too simple or too supersymmetric vacua tend to have some exact moduli but the most generic SUSY-breaking stationary point has no moduli left. People have found "minor" deformations of SUSY-preserving vacua that are nevertheless fully stabilized.

This post imported from StackExchange Physics at 2015-03-23 11:10 (UTC), posted by SE-user Luboš Motl

Comments

Thank you, This is exactly what I was looking for.

This post imported from StackExchange Physics at 2015-03-23 11:10 (UTC), posted by SE-user Prathyush