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The leftover term multiplying the in nitesimal a is our conserved current. Being that we started with a translation of space-time coordinates, we identify this as the momentum: P = T X With Example 7.3 in mind, we can easily nd the conserved supercurrent, which is the conserved current associated with the supersymmetry transformation. Let s just grind it out. Starting with L = T /2 ( X X i ) , we have T 2 ( X ) X i ( ) i ( ) 2 T = 2 ( ) X + ( X ) ( X ) 2 T = 2 ( ) X ( X ) ( X ) 2 = T ( ) X ( X ) = T ( ) X ( X ) ( X ) = T ( X ) X ( X ) + ( X ) = T ( X ) ( X ) The rst term is a total derivative, so it does not contribute to the variation of the action. So we identify the conserved current with the second term. It is taken to be
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The next item of interest in our description of strings with worldsheet supersymmetry is the derivation of the energy-momentum tensor. The energy-momentum tensor is associated with translation symmetry on the worldsheet. Consider an in nitesimal translation which is used to vary the worldsheet coordinates as
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CHAPTER 7 RNS Superstrings
We can write the change of the bosonic elds X by basically writing down their Taylor expansion: X X + X A similar relation holds for the fermionic elds: (7.14)
+
(7.15)
With this in mind, we again follow the Noether procedure. Vary the action as if depended on the worldsheet coordinates, and look for terms multiplied by . At the end we consider to be constant so that term vanishes from the action the term which multiplies will be the energy-momentum tensor that we seek. We proceed in two parts. Let s take a look at the fermionic part of the lagrangian rst. We have i LF = 2 Using Eq. (7.15), we vary this term as follows: i i LF = ( ) ( ) 2 2 i i = ( ) ( ) 2 2 Let s apply the product rule and carry out the derivative on the second term: i i ( ) ( ) 2 2 i i i = ( ) 2 2 2 Now, the variation actually takes place as a variation of the action S, so we can integrate by parts. We do this on the last term to move one of the derivatives off . Integration by parts introduces a sign change, so we get i i i ( ) 2 2 2 i i i = ( ) + ( ) 2 2 2 i i i i = ( ) + + 2 2 2 2
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The divergence term is not going to contribute anything, so we drop it. The rst and third terms cancel, leaving us with i LF = 2 This is what we want, because terms that multiply are going to be terms that make up the energy-momentum tensor T . This isn t quite right, because we want it to be symmetric. So, we take i i LF = 4 4 (7.16)
In the Quiz, you will derive an expression for the bosonic part of the energy-momentum tensor. When all is said and done i i T = X X + + ( Trace ) 4 4 (7.17)
The Trace is explicitly removed to ensure that T remains traceless as required for scale invariance. The energy-momentum tensor and supercurrent can be written compactly using worldsheet light-cone coordinates. The energy-momentum tensor has two nonzero components given by i T++ = + X + X + + + + 2 The components of the supercurrent are