# Kalb-Ramond in T-duality

+ 3 like - 0 dislike
39 views

I'm working through a particular problem (Problem 17.4) in Zweibach's "A first course in string theory", because I'm trying to understand how the Kalb-Ramond field affects T-duality.

The question considers compactification on $T^2$ with a constant Kalb-Ramond field $B_{23} = \frac{1}{2 \pi \alpha^{'}}b$ and all other components vanishing. I have found that the coordinates $x^2$ and $x^3$ (the circle coordinates) have conjugate momenta given by $$p_2 = \frac{\dot{x}^2}{\alpha^{'}} - bw_3$$ $$p_3 = \frac{\dot{x}^3}{\alpha^{'}} + bw_2$$

That is, the effect of the Kalb-Ramond field is to kind of 'twist' the momentum with the winding. We note that the momentum is quantised as $p_r = \frac{n_r}{R}$, and the winding is quantised as $w_r = \frac{m_r R}{\alpha{'}}$. Everything is good until the end, when we're asked to show that the mass-squared operator takes the following form:

$$M^2 = \left( \frac{n_2}{R} + b\frac{m_3 R}{\alpha^{'}}\right)^2 + \left( \frac{n_3}{R} - b\frac{m_2 R}{\alpha^{'}}\right)^2 + \left(\frac{m_2 R}{\alpha^{'}}\right)^2 + \left(\frac{m_3 R}{\alpha^{'}}\right)^2 + \frac{2}{\alpha^{'}}\left(N^{\perp} + \bar{N}^{\perp} - 2\right)$$

This is the part I don't understand. This has the form $M^2 = (\dot{x}^2)^2 +(\dot{x}^3)^2 + w_2^2 + w_3^2 + ...$

whereas from the case of the single circle compactification, I would expect it would be something like $p_2^2 + p_3^2 + w_2^2 +w_3^2+...$ Indeed, when I attempted the calculation that was exactly what I got. This was my attempt at the calculation:

\begin{align} M^2 &= -p^2 \\ &= 2p^+p^- - p^i p^i \\ &= \frac{2}{\alpha^{'}}(L^{\perp}_0 + \bar{L}^{\perp}_0 - 2) - p^i p^i \end{align} For compact directions, the $L^{\perp}_0$ operator picks up a factor of $\frac{1}{2}\alpha_0 \alpha_0 = \frac{\alpha^{'}}{4}(p-w)^2$ for each compact direction, and the $\bar{L}^{\perp}_0$ operator picks up a factor of $\frac{1}{2}\bar{\alpha}_0 \bar{\alpha}_0 = \frac{\alpha^{'}}{4}(p+w)^2$, as shown in Section 17.5 of the book. That is,
$$L^{\perp}_0 = \frac{\alpha^{'}}{4}p^i p^i +\frac{\alpha^{'}}{4}(p_2-w_2)^2 +\frac{\alpha^{'}}{4}(p_3-w_3)^2 + N^{\perp}$$ $$\bar{L}^{\perp}_0 = \frac{\alpha^{'}}{4}p^i p^i +\frac{\alpha^{'}}{4}(p_2+w_2)^2 +\frac{\alpha^{'}}{4}(p_3+w_3)^2 + N^{\perp}$$ With this in hand, completing the computation gives \begin{align} M^2 &= p_2^2 +p_3^2 + w_2^2 + w_3^2 + \frac{2}{\alpha^{'}}(N^{\perp} + \bar{N}^{\perp} - 2) \end{align}

As I expected. While the winding number terms clearly match up with what we are required to show, the momenta don't match up.

Why does the velocity appear in the mass-squared operator rather then the momentum? Have I misinterpreted the role of the various momenta here?

Here is the entire question for completeness:

This post imported from StackExchange Physics at 2016-02-04 18:32 (UTC), posted by SE-user Mark B
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.