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visual basic print barcode label p + i in Java
p + i Quick Response Code Reader In Java Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications. Printing QR Code In Java Using Barcode generator for Java Control to generate, create Quick Response Code image in Java applications. n in e cos(n ) 2 n 0 n
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(2.56) where n is an integer. In addition, periodicity enforces the condition that p = p (2.57) for closed strings. We saw that if this condition was satis ed in the case of open strings, there was no winding of the string permitted. In the case of closed strings, however, the situation is a little bit more involved if we allow for the possibility where the ambient spacetime includes a compact extra dimension (then p = p does not hold). Then, we can consider, for example, the situation where the closed string is compacti ed on a circle of radius R. Using Eq. (2.57), we sum Eqs. (2.55) and (2.56) to obtain the complete solution, focusing on the momentum term and imposing the periodicity condition of Eq. (2.54). This gives the total solution which can be written as X ( , + 2 R) = X ( , ) + 2 RW (2.58) CHAPTER 2 Equations of Motion
We call W the winding number, which literally tells us how many times the string has wound around the compact dimension (so W must be an integer). As we let + 2 , notice that the momentum terms change as p ( + + ) +
p ( ) = p ( + ) +
p ( ) +
2 s ( p p ) 2 That is, the total solution changes as X ( , + ) = X ( , ) +
2 s ( p p ) 2 (2.59) Therefore, we call ( p p ) the winding contribution.
Open Strings with Fixed Endpoints
Finally, we consider open strings with xed endpoints. The boundary condition is X
=0 (2.60) (Dirichlet boundary condition). Using Eqs. (2.48) and (2.49), we have
X L ( , = 0 ) = p + p +
2 s k 0
ik k
(2.61) X R ( , = 0 ) = 2 s k 0
ik k
(2.62) So, Eq. (2.60) implies that p + p = 0 p = p and (2.63) k + k = 0
(2.64) If a dimension is noncompact for an open string, then p = p . To simultaneously satisfy Eq. (2.63), the total momentum of the string must vanish. In the next example, we consider the case where both endpoints are xed. String Theory Demysti ed
EXAMPLE 2.4 What is the length of a string that has both endpoints xed SOLUTION If both endpoints are xed, then we must also satisfy the boundary condition X L ( , = ) + X R ( , = ) = 0 In this case, we have
X L ( , = ) = p +
2 s k 0
ik ( + ) k
X R ( , = ) = p +
2 s k 0
ik ( ) k
The boundary condition can only be satis ed if k is an integer. The overall solution in this case can be written as X =x + n in p 2 s e sin( n ) n 0 n
(2.65) Here we applied the conditions in Eqs. (2.63) and (2.64). This expression includes the winding term w= (2.66) Now let s compute the string coordinates at the endpoints. We have X ( , 0) = x X ( , ) = x + w Hence, the length of the string is X ( , ) X ( , 0) = w CHAPTER 2 Equations of Motion
Poisson Brackets
In going from ordinary classical mechanics to quantum theory, we follow Dirac and use the correspondence between the Poisson brackets and commutators. In string theory, we consider equal Poisson brackets as our starting point when we quantize the modes on the strings. Later, we will discuss hamiltonians and the Virasoro algebra, an important concept in string theory, but for now we simply introduce the important Poisson brackets which will allow us to quantize the string. First we note that {X ( , ), X ( , )} = or, in terms of momentum {P ( , ), X ( , )} = ( ) (2.68) 1 ( ) T (2.67)

