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String Theory Demystified

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CHAPTER 8 Compactifi cation and T-Duality 157<br />

Similarly, using L 0<br />

= 1 together with Eq. (8.17) we obtain<br />

′<br />

= ′ α 2 α 25 25<br />

m p p + 2N −2<br />

2 2<br />

L L L (8.21)<br />

Now using Eq. (8.8) together with Eqs. (8.9) and (8.10) we can write<br />

p<br />

25<br />

= L<br />

Which of course allows us to compute<br />

nR K<br />

′ R<br />

+<br />

α<br />

2 2<br />

25 nR K nR K<br />

pL<br />

R<br />

R<br />

2<br />

( ) =<br />

′ +<br />

⎛ ⎞<br />

⎝<br />

⎜<br />

⎠<br />

⎟ =<br />

⎛ ⎞ ⎛ ⎞<br />

⎝<br />

⎜<br />

′ ⎠<br />

⎟ +<br />

α α ⎝<br />

⎜<br />

⎠<br />

⎟<br />

25<br />

and similarly p = ( K/ R) − ( nR/<br />

α ′ ) so that<br />

R<br />

2 2<br />

25 K nR nR K<br />

pR<br />

R<br />

R<br />

2 ⎛ ⎞<br />

( ) = −<br />

⎝<br />

⎜<br />

′ ⎠<br />

⎟ =<br />

⎛ ⎞ ⎛ ⎞<br />

⎝<br />

⎜<br />

′ ⎠<br />

⎟ +<br />

α α ⎝<br />

⎜<br />

⎠<br />

⎟<br />

This allows us to obtain the sum and difference formulas:<br />

2<br />

2<br />

+ 2<br />

− 2<br />

(8.22)<br />

nK<br />

α ′<br />

(8.23)<br />

nK<br />

α ′<br />

(8.24)<br />

25 nR K<br />

p p<br />

L R<br />

R<br />

2<br />

25 2<br />

2 2<br />

⎡⎛<br />

⎞ ⎛ ⎞ ⎤<br />

( ) + ( ) = 2 ⎢<br />

⎝<br />

⎜<br />

′ ⎠<br />

⎟ +<br />

⎝<br />

⎜<br />

⎠<br />

⎟ ⎥<br />

⎣⎢<br />

α<br />

⎦⎥<br />

(8.25)<br />

25 nK<br />

p p<br />

L R<br />

2<br />

25 2<br />

( ) − ( ) = 4<br />

α ′<br />

(8.26)<br />

Using Eqs. (8.25) and (8.26) we can add Eqs. (8.20) and (8.21) to obtain<br />

2 2<br />

′ =<br />

⎛ ⎞<br />

⎝<br />

⎜<br />

′ ⎠<br />

⎟ + ⎛<br />

2 nR K<br />

α m<br />

⎞<br />

⎝<br />

⎜<br />

⎠<br />

⎟ + 2( NR + NL)<br />

−4<br />

(8.27)<br />

α R<br />

and subtracting Eqs. (8.21) from (8.20) gives<br />

N − N = nK<br />

(8.28)<br />

R L<br />

So notice we have extra terms in the formulas for mass [Eq. (8.27)] and the level<br />

matching condition [Eq. (8.28)] as compared to the formulas introduced for the

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