Rene Meyer
Apr27-04, 10:35 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello,\n\nI tried to derive the formula for the world-sheet supersymmetric\ncurrent (4.1.13) in GSW, p. 189. For this, I assumed, that the spinor\nepsilon is not a constant, and calculated the variation of the action:\n\nThe variations are (I don\'t write space-time indices for simplicity)\n\nx\' - x = deltax = epsbar psi\ndeltapsi = -i rho^a d_a x eps\ndeltapsibar = i epsbar rho^ad_a x\n\nPutting everything in the action, and calculating to first order of\neps:\n\ndelta S = -1/2pi Int d^2Xi [-2 (d_a epsbar) psi d^a x\n-2 epsbar (d_apsi)(d^ax)\n-epsbar rho^b rho^a (d_bx) (d_a psi)\n+psibar rho^a rho^b (d_ad_bx) eps\n+psibar rho^a rho^b (d_bx)(d_aeps)]\n\nThe fourth term vanishes by the equations of motion for the field x,\nd_ad_b x = 0.\n\nNow I can combine the first, second and third term to\n(d_aepsbar)rho^brho^a(d_bx)psi, by first using {rho^a, rho^b} = -2n^ab\non the second and third term, and then integrating by parts,\n\nepsbar rho^a rho^b (d_bx)(d_apsi) = d_a(epsbar rho^a rho^b (d_bx) psi)\n- (d_a epsbar) rho^a rho^b (d_bx) psi\n- epsbar rho^a rho^b (d_ad_bx) psi\n\nwith the first term here being a total differential, thus vanishing,\nand the third again vanishing because of the equations of motion. Thus\nI get\n\ndelta S = -1/2pi Int d^2Xi [ (d_a epsbar) rho^b rho^a (d_bx)psi\n+ Psibar rho^a rho^b (d_bx)(d_aeps)]\n\nBut how can I get the second term vanishing? And what about the\nnormalization? GSW state in (4.1.12) that the factor should be 2/pi,\nthus my sign is wrong, and four times to small. Did I make any mistake\nin the above calculation? Hope someone could help me with this.\n\nRené.\n\n--\nRené Meyer\nStudent of Physics & Mathematics\nZhejiang University, Hangzhou, China\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,
I tried to derive the formula for the world-sheet supersymmetric
current (4.1.13) in GSW, p. 189. For this, I assumed, that the spinor
\epsilon is not a constant, and calculated the variation of the action:
The variations are (I don't write space-time indices for simplicity)
x' - x =[/itex] deltax = epsbar \psi
deltapsi = -i \rho^a d_a x eps
deltapsibar = i epsbar \rho^ad_a x
Putting everything in the action, and calculating to first order of
eps:
\delta S = -1/2pi \Int d^{2Xi} [-2 (d_a epsbar) \psi d^a x
-2 epsbar (d_{apsi})(d^{ax})
-epsbar \rho^b \rho^a (d_{bx}) (d_a \psi)
+psibar \rho^a \rho^b (d_{ad_bx}) eps
+psibar \rho^a \rho^b (d_{bx})(d_{aeps})]
The fourth term vanishes by the equations of motion for the field x,
d_{ad_b} x = .
Now I can combine the first, second and third term to
(d_{aepsbar})\rho^brho^a(d_{bx})\psi, by first using {\rho^a, \rho^b} = -2n^ab
on the second and third term, and then integrating by parts,
epsbar \rho^a \rho^b (d_{bx})(d_{apsi}) = d_a(epsbar \rho^a \rho^b (d_{bx}) \psi)- (d_a epsbar) \rho^a \rho^b (d_{bx}) \psi
- epsbar \rho^a \rho^b (d_{ad_bx}) \psi
with the first term here being a total differential, thus vanishing,
and the third again vanishing because of the equations of motion. Thus
I get
\delta S = -1/2pi \Int d^{2Xi} [ (d_a epsbar) \rho^b \rho^a (d_{bx})\psi
+ Psibar [itex]\rho^a \rho^b (d_{bx})(d_{aeps})]
But how can I get the second term vanishing? And what about the
normalization? GSW state in (4.1.12) that the factor should be 2/\pi,
thus my sign is wrong, and four times to small. Did I make any mistake
in the above calculation? Hope someone could help me with this.
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
I tried to derive the formula for the world-sheet supersymmetric
current (4.1.13) in GSW, p. 189. For this, I assumed, that the spinor
\epsilon is not a constant, and calculated the variation of the action:
The variations are (I don't write space-time indices for simplicity)
x' - x =[/itex] deltax = epsbar \psi
deltapsi = -i \rho^a d_a x eps
deltapsibar = i epsbar \rho^ad_a x
Putting everything in the action, and calculating to first order of
eps:
\delta S = -1/2pi \Int d^{2Xi} [-2 (d_a epsbar) \psi d^a x
-2 epsbar (d_{apsi})(d^{ax})
-epsbar \rho^b \rho^a (d_{bx}) (d_a \psi)
+psibar \rho^a \rho^b (d_{ad_bx}) eps
+psibar \rho^a \rho^b (d_{bx})(d_{aeps})]
The fourth term vanishes by the equations of motion for the field x,
d_{ad_b} x = .
Now I can combine the first, second and third term to
(d_{aepsbar})\rho^brho^a(d_{bx})\psi, by first using {\rho^a, \rho^b} = -2n^ab
on the second and third term, and then integrating by parts,
epsbar \rho^a \rho^b (d_{bx})(d_{apsi}) = d_a(epsbar \rho^a \rho^b (d_{bx}) \psi)- (d_a epsbar) \rho^a \rho^b (d_{bx}) \psi
- epsbar \rho^a \rho^b (d_{ad_bx}) \psi
with the first term here being a total differential, thus vanishing,
and the third again vanishing because of the equations of motion. Thus
I get
\delta S = -1/2pi \Int d^{2Xi} [ (d_a epsbar) \rho^b \rho^a (d_{bx})\psi
+ Psibar [itex]\rho^a \rho^b (d_{bx})(d_{aeps})]
But how can I get the second term vanishing? And what about the
normalization? GSW state in (4.1.12) that the factor should be 2/\pi,
thus my sign is wrong, and four times to small. Did I make any mistake
in the above calculation? Hope someone could help me with this.
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China