T-Duality and the String Length

Classical Mechanics Level 2

Relativistic quantum strings wrapping around a compact dimension have a mass dependent on the winding $w$ and the momentum $p$ , which in turn depend on the radius of the compact dimension $R$ . The mass-squared of certain low-energy states can resultantly be written in terms of the radius as:

$M^2 (R) = \frac{1}{R^2} + \frac{R^2}{\alpha^{\prime 2}} - \frac{2}{\alpha^{\prime}}.$

Find the radius $R$ at which these states are massless, i.e. $M^2 = 0$ .

Bonus: Can you figure out why this is called the self-dual radius ?

\frac{2}{\alpha^{\prime}}

\alpha^{\prime}

\sqrt{\alpha^{\prime}}

\frac{1}{\alpha^{\prime 2}}

1 solution

Matt DeCross
Jan 23, 2016

Setting the mass-squared equal to zero yields the equation:

$M^2(R) = 0 = \frac{1}{R^2} + \frac{R^2}{\alpha^{\prime 2}} - \frac{2}{\alpha^{\prime}}.$

Solving for $R$ by inspection or by factoring the quartic yields $R = \sqrt{\alpha^{\prime}}$ .

T-duality relates spacetimes with compact dimensions of radius $R$ with dual spacetimes that have compact dimensions of radius $\frac{\alpha^{\prime}}{R}$ . At radius $R = \sqrt{\alpha^{\prime}}$ , the radii of the original spacetime and the T-dual spacetime are equal, hence the name self-dual radius .

T-Duality and the String Length

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