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J + = + + X J = X
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i T = X X + 2
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(7.18)
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(7.19)
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The equations of motion for the fermion elds are
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+ = + = 0
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(7.20)
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Together with the equations of motion of the boson elds + X = 0 (7.21)
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CHAPTER 7 RNS Superstrings
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We obtain conservation laws for the energy-momentum tensor: T++ = +T = 0
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(7.22)
EXAMPLE 7.4 Show that the equations of motion for the fermion and boson elds lead to conservation of the supercurrent. SOLUTION We start with J + and consider the derivative J +. We have:
J + = ( + + X ) = ( + ) + X + + ( + X )
=0 The result was readily obtained using Eqs. (7.20) and (7.21). Now taking J , we obtain a second conservation equation by calculating + J which gives
+ J = + ( X )
= + X + + X
= ( + X ) = 0 EXAMPLE 7.5 Show that T++ = 0. SOLUTION Using T++ = + X + X + i /2 + + + , we nd i T++ = + X + X + + + + 2 = ( + X ) + X + + X ( + X ) + i i = + + + = + + + = 0 2 2 To obtain this result, we applied Eqs. (7.20) and (7.21) together with the commutativity of partial derivatives. i i + 2 + + + 2 + + +
String Theory Demysti ed
Mode Expansions and Boundary Conditions
The nal step in putting together the classical physics of the RNS superstring follows the program used in the bosonic case we need to apply boundary conditions and write down the mode expansions. Speci cally, we need to apply boundary conditions for the fermionic elds. It is simplest to continue working in light-cone coordinates and vary the fermionic part of the action. Before doing this, it can be helpful to review some elementary calculus. Recall integration by parts:
f ( x)
b df dg dx = fg b g( x ) dx a a dx dx
The product fg is called the boundary term. When we vary the fermionic action, we are going to obtain boundary terms for the elds , so we need to specify boundary conditions so that the variation in the action vanishes. The fermionic part of the action in light-cone coordinates, modulo a few constants and ignoring the spacetime index is SF = d 2 ( + + + + ) (7.23)
For simplicity, let s consider one piece of this expression and vary it. We obtain
d 2 + + = d 2 [ + + + + ( + )]
Following the usual procedure applied in eld theory, we want to move the derivative off the + term. This can be done using integration by parts. When this is done, we pick up a boundary term:
d
( + ) = d + +
= =0
d 2 + +
A similar expression arises from the variation of the other term. All together, the boundary terms obtained by varying the action are
SF = d ( + + )
=
( + + )
=0
(7.24)
CHAPTER 7 RNS Superstrings
OPEN STRING BOUNDARY CONDITIONS
When varying the action, the boundary terms must vanish in order to maintain Lorentz invariance. In the case of open string, the boundary terms = 0 and = must both vanish independently. We can obtain
+ + = 0
at = 0 if we take
+ ( 0, ) = ( 0, )
(7.25)
Now in general, + = will make the boundary terms vanish, but typical convention is to x the boundary condition at = 0 using Eq. (7.25). This leaves the choice of sign at = ambiguous. Depending on the sign we choose, we obtain two different boundary conditions. Ramond or R boundary conditions are given by the choice
+ ( , ) = ( , )
(Ramond)
(7.26)
The other choice we can make is known as Neveau-Schwarz or NS boundary conditions:
+ ( , ) = ( , )
(Neveau-Schwarz)
(7.27)
We often refer to the boundary conditions chosen as the sector. The choice of boundary conditions has dramatic consequences. In particular The R sector gives rise to string states that are space-time fermions. The NS sector gives rise to string states that are space-time bosons.
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